UC-NRLF 


B   M   501   7Db 


o 


m  ■ 


l%!w^  ■«■■■■;■  • 


till 


•:*■.:;  .>i!K; 


m 


m^ 


iiiSiiii: 


^i/<, 


AN 


ELEMENTARY  COURSE 


IN 


Descriptive  Geometry. 


SOLOMON   WOOLF,  A.M., 

1 1 
-Professor  of  Descriptive  Geometry  and  Drawir.j  in  I  he  College  of  the  City  of  New    York. 


THIRD   EDITIOS. 


SECOM)  TBOrSAXP. 


NEW    YORK: 

JOHN    WILEY    &    SONS. 

LONDON : 
CHAPMAN   &    HALL,  Ltd. 


'■"•'EfiAL 


Copyright,  1888, 
By  JOHN  WILEY  &  SONS. 


TABLE   OF   CONTENTS. 


CHAPTER   I. 
GENERAL   CONSIDERATIONS. 
Definition: — Planes  of  Projection — Dihedral  angles  (Fig.  i),        .  ,  ,  .1 

The  point  in  space : — Orthographic  projection. — Projecting  lines  and  ordinates — Rota- 
tion of  planes  into  drawing  plane  (Figs.  2,  3),      .  .  .  .  .  .2 

Change  of  Ground  line : — Notation  of  point — Positions  of  point  in  space  (Figs.  4,  5),      .       5 
Problem. — ^To  find  the  projections  of  a  point,  its  distances  from  the  planes  of  projection 

being  given.     Various  examples  (Fig.  6),  .  .  .  .  .  .  .6 

Problem. — The  primitive  projections  of  a  point  being  given,  to  find  its  new  vertical  pro- 
jection upon  any  new  vertical  plane  (Fig.  8),        .  .  •  .  .  .7 

CHAPTER   II. 
THE   LINE. 

The  line  in  space  and  its  projections  (Fig.  9), .  .  .  .  ,  ,  .8 

Problem. — The  projection  of  two  points  being  given,  to  find  the  projections  of  the  line 

which  they  determine  (Figs.  12-13),  •  •  •  •  •  •  '9 

Problem. — Given  the  projections  of  two  points,  to  find  the  projections  of  a  third  point 

of  the  line  which  they  determine  (Figs.  14-18)..  .  .  .  .  .10 

Problem. — A  line  being  given,  to  find  upon  that  line  (i)  the  vertical  projection  of  any 

given  point;  (2)  the  horizontal  projection  of  any  given  point  (Figs.  19-20).  .  .     il 

Problem. — A  line  being  given,  to  find  upon  that  line  (i)  the  point  whose  distance  from 

//  is  a  minimum  ;  (2)  the  point  whose  distance  from  K  is  a  minimum  (Figs.  21-23),     11 
Problem. — To  find  the  traces  or  piercing-points  of  a  line  the  projections  of  which  are 

given.  ...........      12 

Lines  parallel  to  both  planes  of  projection  (Fig.  24),  .  .  .  •     '3 

Graphical  characteristics  of  lines,        .  .  .  .  .  .  .  -13 

Lmes  parallel  to  t>««^  plane  of  projection  (Figs.  25-28),         .  .  .  .  •     '3 

Lines  parallel  to //<v///£-r  plane  of  projection  (Figs.  29-30),    .  .  .  .  •     '5 

Problem.— To  find  the  projections  of  a  right  line  parallel  to  the  ground-line  (Fig.  31),    .      16 
Problem.— To  find  the  projections  of  a  right  line  perpendicular  to  either  coordinate 

plane  (Fig.  32,  i).  .  .  .  .  .  .  .  .  ,  .16 

Problem. — To  find  the  projections  of  a  line  parallel  to  one  plane  of  projection  and  in- 
clining to  the  other  (Fig.  32,  2),    .  .  .  .  .  ,  .  ,16 

Problem.  — To  find  the  projections  of  any  line,  its  piercing-points  being  given  (Fig.  33),  .     17 
Problem.— A.  line  being  given  by  its  projections,  to  find  a  point  of  that  line  whose  dis- 
tances from  the  coordinate  planes  shall  be  in  a  given  ratio  (Fig.  34),   .  .  •      I7 
Problem.— A.  Ime  being  given  by  its  projections,  to  find  a  point  of  that  line  whose  dis- 
tances from  the  coordinate  planes  shall  be  equal  (Fig.  35),        .             ,             .             .      iS 


IV  TABLE    OF  CONTENTS. 

CHAPTER    III. 
LINES,    PLANE    FIGURES   AND    PLANES. 

PAGE 

Parallel  lines.    ...-•••••••     i9 

Problem.— i:^xow^\\  a  given  point  to  pass  a  line  parallel  to  a  given  line  (Figs.  36.  37).        .      19 
Intersecting  lines,  ..........     20 

Problem.— 'Yo  analyze  the  positions  of  lines  in  space  whose  projections  are  given,  .     20 

Problem. — To  determine  the  relative  position  of  two  lines  when  either  or  both  lie  in  a 

plane  perpendicular  to  GL  (Figs.  41,  42),  .  .  .  .  .  .21 

Problem.— 10  determine  the  projections  of  any  line  connecting  two  given  lines  which 

intersect  (Fig.  43)'  •  •  •  •  •  •  •  •  .21 

Problem.— To  determine  the  relative  position  of  two  lines  whose  projections  do  not  in- 
tersect withm  the  limits  of  the  drawing  (Figs.  44,  45),  .  .  .  .  .22 

Problem.— 1^0  draw  the  projections  of  a  line  passing  through  a  given  point  and  inter- 
secting a  given  line,  when  either  projection  of  the  point  of  intersection  lies  beyond 
the  limits  of  the  drawing  (Fig.  46),  .  .  .  .  .  .  .22 

Plane  figures  (Figs.  47,  48),      .........     23 

Problem.— To  find  the  projections  of  any  plane  figure  parallel   to  the  vertical  plane 

(Fig.  49) 23 

Problem. — To  find  the  projections  of  a  circle  parallel  to  the  horizontal  plane  (Fig.  50),    .     23 
Problem. — To  find  the  projections  of  a  rectangle  the  plane  of  which   is  perpendicular  to 

F and  inclines  to //at  any  given  angle  (Fig.  51).  .  .  .  .  .23 

Problem.— To  determine  the  projections  of  a  circle  the  plane  pf  which  is  perpendicular 

to //"and  inclines  to  F  at  any  given  angle  (Fig.  52),        .  .  .  .  .24 

Planes  (Figs.  53-59),         ........••     24 

Horizontals,  verticals,  and  lines  of  greatest  declivity  (Figs.  60-62).  .  .  .  .26 

Notation  of  the  plane,      .  .  .  •  .  .  .  •  •  -27 

Problem.— To  analyze  the  position  of  a  plane  given  by  its  traces  (Fig.  63),  .  .     27 

Problem. — To  pass  through  a  line  in  space  a  plane  perpendicular  to  either  coordinate 

plane  (Figs.  64, 65),  .  .  .  ...  .  .  •  •  .27 

Problem.— Given  three  points  not  in  the  same  right  line,  to  find  any  line  of  the  plane 

which  they  determine  (Figs.  66-71),  .  .  .  .  .  .  -27 

Problem. — Given  the  projections  of  three  points. not  in  the  same  right  line,  to  find  any 

point  of  the  plane  which  they  determine, .  .  .  .  .  .  -29 

Problem. — Given  one  projection  of  a  line  lying  in  a  given  plane,  to  find  the  other  pro- 
jection (Figs.  72-74),  .  .  .  .  .  .  .  .  -29 

'Problem. — Given  the  traces  of  a  plane,  to  find   the  projections  (i)  of  any  horizontal, 

(2)  of  any  vertical  (Figs.  75, 76),      .  .  .  .  .  .  .  -3° 

Problem. — The  plane  being  given  by  three  points  not  in  the  same  right  line,  to  find  a 

vertical  or  a  horizontal,     .  .  .  .  .  .  .  .  -3° 

Problem. — Given  one  trace  of  a  plane  and  the  projections  of  any  point  in  that  plane,  to 

find  the  other  trace  (Figs.  T],  78),  .  .  .  .  .  .  .30 

Problem. — To  find  the  traces  of  a  plane  passing  through  two  intersecting  lines  (Fig.  79),     30 
Problem.— To  pass  a  plane  through  three  given  points  (Fig.  80),     .  .  .  -3' 

Problem. — To  pass  a  plane  through  a  given  point  and  line,  .  .  .  -31 

Problem. — Given  a  plane  by  its  traces,  to  determine  a  new  vertical  trace  by  a  change  of 
ground-line  (Figs.  81-83),  .  .  .  .  .  .  .  •     3' 

CHAPTER    IV. 

I.   PARALLEL   AND    INTERSECTING    PLANES. 

Parallel  planes,     ...........     33 

Problem.— To  pass  through  a  given  point  a  plane  parallel  to  a  given  plane  (Fig.  84),         .     33 
Problem. — Through  a  given  line  to  pass  a  plane  parallel  to  a  second  line  (Fig.  85),  .     33 

Problem — Through  a  given  point  to  pass  a  plane  parallel  to  two  given  lines,         .  .     34 


TABLE    OF  CONTENTS. 


Intersecting  planes,  .......... 

Problem : — To  find  the  line  of  intersection   between  two  planes,  given  by  their  traces 

(Figs.  86-IOI),  .......... 

ProbUm. — To  find  the  line  of  intersection  between  two  planes  when  the  traces  do  not 

intersect  within  the  limits  of  the  drawing  (Figs.  102-103),         .  .  .  . 

Positions  of  three  intersecting  planes  and  their  lines  of  intersection, 

II.  THE  LINE   AND    PLANE. 
Probhin.—'Yo  find  the  piercing-point  of  a  line  on  a  given  plane  (Figs.  104-111),    . 
Thtoroit. — A  right  line  perpendicular  to  a  plane  in  space  gives  projections  which  are 

respectively  perpendicular  to  the  traces  (Figs.  112,  113),  .  .  .  . 

Problem. — Through  a  given  point  in  space  to  pass  a  plane  perpendicular  to  a  given  line 

(Fig.  114).       ........... 

Problem. — Through  a  given  line  to  pass  a  plane  perpendicular  to  a  given  plane  ( Fig.  1 1 5), 
Problem. — To  pass  a  plane  through  a  given  point  and  parallel  to  two  given  lines  (Fig. 

116), 

CHAPTER   V. 
I.  GEOMETRICAL   SURFACES. 
Generation  and  division  of  geometrical  surfaces,        ...... 

II.   PLANE   SURFACES. 
The  prism  and  the  pyramid,     ...... 

Problem. — To  find  the  projections  of  a  parallelopiped  (Fig.  117),    . 
Projections  of  the  cube  and  prism  (Figs.  118.  1 19),   . 
Problem.— 'Xo  project  a  right  pyramid  (Figs.  120,  121), 
Regular  poi\  hedrons.   ....... 

Problem. — To  determine  the  projections  of  the  octahedron  (Fig.  122), 
Problem. — To  determine  the  projections  of  the  icosahedron  (Fig.  123), 
Consideration  of  the  seen  and  hidden  lines  in  projections,  . 
Rules  to  be  obser\^ed  (Figs.  126,  127),  .... 

Problem. — To  project  an  oblique  pyramid  (Fig.  128), 
Problem. — To  project  an  oblique  prism  (Fig.  129),     . 

III.  SINGLE-CURVED   SURFACES. 
The  cylinder  and  cone.  ...... 

Proble/n. — To  project  a  right  cone  or  right  cylinder  (Figs.  130,  131,) 
Problem. — To  project  an  oblique  cone  or  oblique  cylinder  (Figs.  132,  133) 

IV.  WARPED    SURFACES. 
The  hyperboloid  of  one  nappe  (Fig.   134),  the  warped  plane  (Fig.    135).  right  conoid 

«F'g-  '36) 

The  helicoids  and  helix  (Fig.  137).      ..... 

Problem. — To  project  a  warped  helicoid  (Fig.  139).  . 

V.   DOUBLE-CURVED   SURFACES  AND   SURFACES   OF    REVOLUTION. 
A.xes,  meridian  lines  and  planes,  and  parallels. 
Ellipsoid,  hyperboloid,  paraboloid,  and  sphere, 
Triaxial  ellipsoid,  hyperboloid  of  one  nappe,  and  elliptic  paraboloid  (Figs.  140-142). 

CHAPTER    VI. 
SUPPLEMENTARY    PLANES   AND    PROJECTIONS. 
Nature  and  use  of  supplementary  planes.  ....... 

Supplementary'  projections  (Figs.  143-147),      ....... 

Problem. — The  traces   of  two  planes  being  given,  to  find  their  line  of  intersection  by 
means  of  a  profile  supplementar)-  plane  (Fig.  148),  ..... 


PAGE 

34 


34 
36 

n 

38 

40 

40 
40 


42 


Vi  TABLE    OF  CONTENTS. 

PAGB 

Problem.— 1o  determine  the  oblique  supplementary  projection  of  a  circle,  its  primitive 

projections  being  given  (Fig.  149)-  •  •  •  •  '  *       .     '     ^'^ 

Problem.— To  find  the  primitive  projections  of  a  right  prism,  its  supplementary  projec- 
tion being  given  (Fig.  150),  .  .  •  •  •  •  •  -5 

CHAPTER   VII. 

CHANGE   OF   POSITION    BY    ROTATION    AND    RABATTEMENT. 
The  method  of  rotation  explained  (Fig.  151),  .  .  •  •  •  •  '59 

I.    ROTATION    OF   THE    POINT. 

Problem.— 10  revolve  a  point  into  either  coordinate  plane  around  an  axis  lying  in  that 

plane  (Figs.  152,153) •  .*,."..     ' 

Problem.— To  revolve  a  point  into  either  coordinate  plane  around  an  axis  lymg  m  that 
plane  when  neither  projection  of  the  point  coincides  with  the  projection  of  the  axis 

(Figs.  154,  155).         •  •  •  •  •  •  •  \        .  * 

Problem.— To  revolve  a  point  through  a  given  angle  around  a  horizontal  axis  (Fig.  156),  .  62 
Problem.— To  revolve  a  point  through  a  given  angle  around  a  vertical  axis  (Fig.  157),  .  63 
Problem.— To  revolve  a  point  through  a  given  angle  around  an  axis  parallel  to  either 

coordinate  plane  (Figs.  158,159),    ....•••• 

II.   ROTATION   OF  THE   LINE. 

Problem.— To  revolve  a  right  line  around  a  vertical  axis  and  through  a  given  angle  (Figs. 

160,  161),         .  .  .  •  •  •  •  •  ■   J.       *      ,      ' 

Problem.— To  revolve  a  right  line  until  it  becomes  parallel  to  either  coordmate  plane 

(Figs.  162,  163). ■     ^\ 

Problem.— To  revolve  a  right  line  parallel  to  the  ground-lme,           .            .            •  -04 

Problem.— To  revolve  a  right  line  until  it  becomes  perpendicular  to  the  vertical  plane,  .     64 

Problem.— To  revolve  a  line  until  it  becomes  perpendicular  to  the  horizontal  plane,  .     64 

III.    ROTATION    OF   THE    PLANE. 

Data  for  the  rotation  of  a  plane,  .  .  •  •  •  'r.-         i-       r  ^'     a 

Problem.— To  revolve  a  plane  around  a  vertical  axis  through  a  given  angle  (Figs.  164,  165),     65 
Probletn  —To  revolve  a  plane  around  an  axis  perpendicular  to  the  vertical  plane  (Figs. 


60 


61 


62 


65 


166,  167) 

IV.    RABATTEMENT. 


Rabattement  and  counter-rotation,        .  .  •  •  •  ■,•"," 

/'r^^/^m.-Rabattement  of  a  plane  around  a  vertical,  the  plane  being  perpendicular  to  ^ 

the  vertical  plane  (Figs.  168,  169),  .  .  ■  •  ■  •  •  '  J^ 

Pr^^/m.— Rabattement  of  a  plane  around  one  of  its  horizontals  (Figs.  170,  171).  •  07 

Proble7n.—T\iXOM^\^  a  given  point  of  a  plane  to  draw  a  rectangle  in  that  plane  (Fig.  172),  6» 

Prtf^/^z/z.— To  construct  a  triangle  on  any  line  of  a  given  plane  (Fig.  173).  •  •  ^9 

CHAPTER   VIII. 

DISTANCES   AND   PERPENDICULARS. 
L    DISTANCE   OF   TWO    POINTS. 

Problem.— To  determine  the  distance  between  two  points  given   by  their  projections 

(Figs.  174-177).  •  •  •  •  •  •  '  '  •  '   /tr- 

Problem.— \5^^ori  a  given  line  to  measure  a  given  distance  from  either  extremity  (tigs. 

178,179) ^' 

II.    DISTANCE   OF    POINT    FROM    LINE. 

Problem.— To  determine  the  perpendicular  between  a  point  and  a  line  given  by  their 
projections  (Figs.  180-183),  .  •  •  ■  •  '        ,•       /c-     ' 

Problem.— Vxom  a  given  point  as  a  vertex,  to  construct  a  triangle  on  a  given  line  (Hgs. 
184,  185),         .  .  .  ■  •  •  •  •  • 


71 


TABLE    OF  CONTENTS.  VU 

III.    DISTAN'CE  OF   POINT    FROM    PLANE, 

l-AGH 

Problem. — To  determine  the  distance  between  a  point  and  a  plane  (Figs.  186-189),  .     74 

Problem. — Through  a  given  point  in  a  line  to  pass  a  plane  perpendicular  to  that  line,      .     76 
Problem. — Through  a  given  point  to  pass  a  plane  perpendicular  to  a  given  plane.  .     76 

Problem.— lo  determine  the  distance  between  two  parallel  planes. .  .  .  .76 

Problem. — To  erect  a  right  prism  standing  on  a  given  platie  (Fig.  190J.       .  .  .76 

Problem.— G\\;&n  a  plane  and  the  rabattement  of  the  base  of  a  pyramid  standing  on  that 

plane  to  determine  the  projections  of  the  pyramid  after  counter-rotation  (Fig.  191),      77 
IV.   SHQRTEST   DISTANCES   BETWEEN   TWO    LINES. 
Perpendiculars  in  direction  (Figs.  192,  193),      .  .  .  .  .  .  .78 

Perpendiculars  in  position  (Fig.  194),    .  .  .  .  .  .  .  .78 

Problem. — To  determine  the  line  measuring*  the  shortest  distance  between  two   right 

lines  not  in  the  same  plane  (Figs.  195-199).  .  •  •  .  .  -79 

Problem. — Given  the  horizontal  projections  of  two  lines  and  the  projections  of  their  line 

of  shortest  distance,  to  determine  the  vertical  projections  (Fig.  200J,    .  .  .81 

Problem. — Given  the  projections  of  a  line  and  the  horizontal  projections  of  a  second  line 

and  the  line  of  shortest  distance,  to  determine  their  vertical  projections  (Fig.  201),  .     82 

CHAPTER   IX. 

ANGLES. 
I.    ANGLES    BETWEEN   RIGHT   LINES. 

Problem. — To  determine  the  angle  between  two  lines  whose  projections  are  given  (Figs. 

202-205),         ...........    83 

/'/-tfM///.— To  determine  the  angles  which  a  line  in  space  makes  with    its  projections 

(Figs.  206,  207).         ..........     84 

Problem.— 1:0  determine  the  angle  between  the  traces  of  a  given  plane  (Fig.  208),  .     84 

Problem.— Ihe.  projections  of  two  intersecting  lines  being  given,  to  bisect  or  otherwise 

divide  their  angles  (Fig.  209),        .  .  .  .  .  .  .  -84 

Problem.— Through  a  given  point  to  pass  a  line  which  intersects  a  given  line  at  any 

angle  (Figs.  211,  212).  .  .  .  .  .  .  .  .  .85 

II.    ANGLES    BETWEEN    LINES   AND    PLANES. 
Problem. — To  determine  the  angle  between  a  line  and  a  plane  (Figs.  213-215),     .  .     86 

Problem. — To  pass  through  a  given  point  a  line  making  any  angle  with  a  plane  (Fig.  216),     86 
Lines  of  greatest  declivity,  .  .  .  .  .  .  •  •  -87 

Problem. — To  determine  a  plane  when  its  line  of  greatest  declivity  is  given  (Figs.  217. 

2.8) 87 

Probhm.—'Vo  determine  the  projections  of  a  line  when  its  inclinations  to  the  coordi- 
nate planes  are  given  (Fig.  219),      .  .  .  .  .  •  •  .88 

III.    ANGLES    BETWEEN   PLANES. 
Measure  of  the  angle  between  planes.  .  .  .  .  .  •  •  .88 

Problem. — To  determine  the  angle  between  two  planes  (Figs.  220-22^),      .  .  .88 

Problem. — To  determine  the  angles  which  a  given  plane  makes  with  the  coordinate  planes 

(Figs.  224,  225),  ..........     90 

Problem.— To  determine  the  planes  which  contain  the  line  of  inter>cction  between  two 

planes  and  which  bisect  their  angles  (Fig.  226),  .  .  •  •  -9° 

Problem. — To  find  the  traces  of  a  plane  when  its  inclination  to  the  coordinate  planes  is 

given  (Figs.  227,  228),         .  .....•.•     9° 

CH.\PTER   X. 

CHANGE   OF    POSITION       "   COMBINED    MOTIONS. 
Problem. — Given  a  triangle,  required  to  find  i  'vlicn  the  axis  is  made  to  in- 

cline to  both  planes  of  projection  (Fig.  2 


Vlll  TABLE   OF  CONTENTS. 

PAGE 

Problem. — To  project  a  cube  in  a  doubly  oblique  position  (Fig.  230),        .  .  '93 

Problem. — To  determine  the  projections  of  an  oblique  hexagonal  pyramid  in  a  doubly 

oblique  position  (Fig.  231),  .  .  .  .  .  .  .  -94 

Problein. — To  determine  the  projections  of  a  right  cylinder  in  a  doubly  oblique  position 

(Fig.  232) 95 

Proble)n.—1o  determine  the  projections  of  a  right  cone  when  the  inclination  of  the  axis 

to  the  coordinate  planes  is  given  (Fig.  233),       .  .  .  .  .  .96 

CHAPTER   XI. 

SECTIONS. 

I,   ELEMENTARY   SECTIONS. 

Character  of  section  as  affected  by  position  of  cutting  plane  and  nature  of  surface,  .    98 

Longitudinal,  meridian  and  perimetrical  sections,       .  .  .  .  .  .98 

Open  and  closed  sections,  .  .  .  .  .  .  .  .  -99 

Elementary  sections  and  breaking-points,        .......  100 

Selection  of  cutting  planes  in  graphical  constructions,  .....   100 

Problem. — Through  a  point  on  the  surface  of'a  cone  to  cut  a  longitudinal  and  a  parallel 

section  (Fig.  234),     .....  .  .  .  .   loi 

Problem. — Through  a  point  on  the  surface  of  an  oblique  cylinder  to  cut  a  longitudinal 

and  a  parallel  section  (Fig.  235),    .  .  .  .  .  .  .  .  loi 

Problem. — To  divide  a  horizontal  prism   into  two  equal   parts  by  a  longitudinal  plane 
which  makes  a  given  angle  with  the  horizontal  plane ;  and  to  remove  the  front  sec- 
tion of  the  surface  (Fig.  236),       .  .  .  .  .  .  .102 

Problem.— To  cut  the  meridians  on  a  surface  of  rotation  (Fig.  237),  .  .  .102 

II.    OBLIQUE  SECTIONS. 

Nature  and  determination  of  oblique  sections,  ......   103 

Problem. — To  intersect  a  given  prism  by  a  plane  perpendicular  to  the  vertical  plane 

(Fig.  238),   .  . 103 

Problem. — To  intersect  an  oblique  pyramid  by  a  plane  perpendicular  to  the  horizontal 

plane  (Fig.  239),     .  .  .  .  .  .  .  .  .  .104 

Problem.— To  hitersect  a  right  cylinder  by  a  given  plane  (Fig.  240),  .  .  .105 

Problem. — To  intersect  a  right  cone  by  a  given  plane  (Figs.  241-243),        .  .  .105 

Problem.— To  intersect  a  surface  of  revolution  by  a  given  plane  (Fig.  244),  .  .   107 

Problem. — To  intersect  a  pyramid  by  a  given  plane  (Fig.  245),  ....   108 

Problem.— To  intersect  an  oblique  cylinder  by  a  plane  at  right  angles  to  the  axis  (Fig. 

246),  .  .  .  .  .  .  .  .  .  .109 

Problem. — To  pass  a  plane  through  a  given  line,  and  to  cut  a  great  circle  on  a  sphere 

(Fig.  247) no 

CHAPTER  XH. 

INTERSECTIONS. 
I.    ELEMENTARY   INTERSECTIONS. 

The  line  of  intersection  as  affected  by  nature  of  surface,  position  and  extension,  .  iii 

Requirements  for  elementary  intersections,  .......   112 

Problem. — To  dispose  a  right  prism  and  right  cylinder  so  as  to  intersect   in  rectilinear 

elements  (Fig.  248),  .  .  •  •  •  •  .  .  .112 

Problem. — To  dispose  two  cones  so  as  to  intersect  in  rectilinear  elements  (Fig.  249),  .  112 
Problem. — To  dispose  a  right  pyramid  and    right  prism  whose  respective  parallels  are 

similar,  so  as  to  intersect  in  a  parallel  (Fig.  250),  .  .  .  .  .112 

Problem. — To  dispose  surfaces  of  revolution  so  as  to  intersect  in  parallels  (Fig.  251),  .  113 
Problem. — To  determine  the  circle  of  intersection  between  two  spheres  (Fig.  252).  .    113 


TABLE   OF  CONTENTS. 
II.    PLANE    LINES   OF    INTERSECTION. 


PACE 


Requirements  for  plane  intersections, .  .  ,  .  .  -114 

Problem. — To  find  the  line  of  intersection  between  two  equal  cylinders  whose  axes  in- 
tersect (Fif(.  253J,     .  .  .  .  .  .  ,  .  .  .115 

Problem. — To  find  the  line  of  intersection  between  two  equal  prisms  whose  axes  inter- 
sect at  right  angles  (Fig.  254),       .  .  .  .  .  .  .  •    "5 

Problem. — To  find  the  line  of  intersection  between  two  equal  pyramids  whose  axes  inter- 
sect at  right  angles  (Fig.  255),       .  .  .  .  .  .  .  .116 

Problem. — To  find  the  line  of  intersection  between  two  equal  cones  (Fig.  256J,     .  •   "7 

III.    INIERSECTIONS   IN    WARPED    LINES. 

Variations  Jn  the  nature  of  the  lines  according  to  the  surfaces,       .  .  .  .118 

Problem. — To  find  the  line  of  intersection  between  a  rigiit  pyramid  and  a  right  prism 

(Fig.  257j.    .  .  .  .  .  .  .  .  .  .  .119 

Problem. — To  find  the  line  of  intersection  between  a  right  prism  and  a  sphere  (Fig.  258),  119 
Problem. — To  find  the  line  of  intersection  between  two  oblique  prisms  (Fig.  259),  .   120 

Problem. — To  find  the  line  of  intersection  between  a  cylinder  and  a  cone  (Fig.  260),  .  121 
Problem. — To  find  the  line  of  intersection  between  a  cylinder  and  a  sphere,  the  sphere 

being  tangent  interiorly  to  the  cylinder  (Fig.  261),         .  ...  .  .    122 

Problem. — To  find  the  intersection  between  two  surfaces  of    revolution  whose  axis  line 

is  a  common  meridian  plane  (Fig.  262),  .  .  .  .  .  .  -1-3 

Problem. — To  find  the  intersection  between  two  surfaces  of  revolution  whose  axes  du 

not  lie  in  the  same  plane  (Fig.  263),  .  .  .  .  .  .  .124 


CHAPTER    XIII. 
TANGENTS  AND    NORMALS. 
I.    GENERAL  CONSIDERATIONS. 

The  tangent,  point  of  contact,  normal  and  normal  plane  (Figs.  264,  265),  .            .126 

The  point  of  inflexion,  cusp  point,  and  angle  of  curves  (Fig.  266),   .  .            .            .126 

Tangent  lines,  planes  and  surfaces,       .            .            .            .            .  .            .            .1-7 

II.  TANGENIS  AND  NORMALS  TO  SURFACES. 

Tangents  to  ruled  surfaces,         .            .            .            .            .            .  .            .            .128 

Tangents  and  normals  to  double-curved  surfaces,       .            .            .  .            .            .129 

Tangents  and  normals  to  surfaces  of  rotation,             .             .             .  .             .             •   '-9 

Problem. — To  pass  a  line  tangent  to  a  section  on  the  cylinder,  the  point  of  contact  being 

given  (Fig.  240),        .            .            .                         .            .            .  .129 

Problem. — To  pass  a  line  tangent  to  a  section   on  the  cone,  the  point  of  contact  being 

^iven  (Fig.  241),        .                          .             .             .             .             .  .             .             .130 

Problem. — To  pass  a  line  tangent  to  a  section  on  a  cylinder  (Fig.  246). 


MO 


Problem. — To  find  a  tangent  to  any  plane  curve  n-  •  -  r  '.     "■  >r  .-trcal  definition,     .  130 

III.    TANGENTS   TO    RUI 

Problem. — To  pass  a  plane  tangent  to  a  right  cyliiu.  section,  and  de- 
termine the  tangent  line  to  that  section  (Fig.  2(  .                          .  130 
Problem. — Through  a  given  point  in  space  to  pass  a  pi  ier  (Fig.  268),  131 
Problem. — To  pass  a  plane  tangent  to  a  cylinder  and  \  •  ?  <  Fig.  269).    .  132 
Problem. — Given  one  projection  of  a  point  on  the  surf  through  that 

point  a  plane  tangent  to  the  cone  (Fig.  270),        .  •             •  ^ll 

Problem. — To  pass  through  a  point  in  .space  a  plane  tai  !    ^.  271).             •  ^l}) 

Problem. — To  pass  a  plane  tangent  to  a  cone  and  parall<  .le  rFigr.  272),          .  133 


TABLE    OF  CONTENTS. 


IV.    SURFACES    OF    REVOLUTION. 

Problem. — Through  a  point  on  a  surface  of  revolution  to  pass  a  plane  tangent  (Fig.  273), 
Problem. — Through  a  given  point  in  space  to  pass  a  plane  tangent  to  a  sphere  at  a  given 

meridian  (Fig.  274),  ......... 

Problem. — To  pass  a  plane  tangent  to  a  surface  of  revolution  and  parallel  to  a  given 

plane  (Fig.  275),        .  .  .  . 

Problem.— 1:0  pass  a  plane  parallel  to  a  given  line,  and  tangent  to  a  surface  of  revolution 

at  a  given  meridian  (Fig.  276),        ........ 

V.   WARPED   SURFACES. 

Problem. — The  axis  and  generatrix  of  a  warped  surface  of  rotation  being  given,  to  pass  a 
plane  tangent  at  a  given  point  of  the  surface  (Fig.  277),  .... 

Problem. — Through  a  given  point  in  space  to  pass  a  plane  tangent  to  a  warped  surface 
of  rotation  at  a  given  meridian  (Fig.  278;,  ...... 


PAGE 

136 
137 

137 
139 


,    VI.   TANGENT   SURFACES   AND   ENVELOPMENT. 

Definitions  and  explanations ;  tangent  juncture,         .  .  .  .  .  .  140 

Proble?tt. — To  circumscribe  a  sphere  by  a  cylinder  whose  axis  is  parallel  to  a  given  line, 

and  determine  the»Iine  of  contact  and  the  horizontal  trace  of  the  cylinder  (Fig.  279),  141 

Problem. — Through  a  given  point  in  space  to  pass  a  plane  tangent  to  a  surface  of  rota- 
tion, the  point  of  contact  being  on  a  given   parallel  (Fig.  280),  .  .  ,142 

Problem. — To  determine  a  similar  tangency  by  means  of  an  auxiliary  sphere  (Fig.  281),    143 


CHAPTER   XIV. 

DEVELOPMENT. 

Explanation  : — Development  of  rectilinear  surfaces. 

Development  of  ruled  surfaces,  ..... 

Problein. — To  develop  a  right  prism  with  oblique  section  (Fig.  282), 

Problem. — To  develop  the  section  of  a  right  cylinder  (Fig.  283), 

Problem. — To  develop  the  section  of  an  oblique  cylinder  (Fig.  284), 

Problem. — To  develop  a  right  regular  pyramid  and  section  (Fig.  285), 

Problem. — To  develop  a  right  cone  of  revolution  and  section  (Fig.  286), 

Problem. — To  develop  the  oblique  pyramid  and  section  (Fig.  287), 

Problem. — To  develop  the  oblique  cone  (Fig.  288), 

Problem. — To  develop  the  icosahedron  (Fig.  289),      .  .  • 


145 
146 
146 

147 
147 
148 

149 
150 

151 
152 


ELEMENTS  OF  DESCRIPTIVE  GEOMETRY. 


CHAPTER  I. 


GENERAL  CONSIDERATIONS. 


1.  In  a  limited  sense,  Descriptive  Geometry  may  be  defined  to  be  a 
conventional  method  of  representing  on  a  plane  objects  which  have  three 
dimensions,  so  as  to  admit  of  an  accurate  determination  of  their  size, 
form  and  position. 

2.  But  position  being  relative,  objects  are  in  respect  of  Tihis  undeter- 
minable unless  referred  to  other  objects,  which,  for  constructive  pur- 
poses, ought  to  be  of  the  simplest  character  and  in  positions  readily 
conceived   and  comprehended. 

In  Descriptive  Geometry  these  latter  objects  are  two  or  more  planes,* 
termed  the  planes  of  projection. 

3.  These  two  planes  intersect  at  right  angles  (Fig.  i),  one  being  in 
the   position  of   an  ordinary    wall   and    the    other  in   the   position   of    the 


n 

I 
L 

^^'<ff 

^ 

IJy^ 

m 

•Jy^ 

IV 

floor.  The  upright  plane,  V,  is  the  vertieal  plane  of  projection,  and  the 
other  plane.  H,  at  right  angles  to  it,  the  horizontal  plane  of  projection. 
The  line   GL  in  which  they  intersect  is  termed  the  ground-line. 

4.  The  planes  thus  arranged  are  divided  by  their  line  of  intersection, 
GL,  into  two  parts  each,  and,  in  turn,  divide  space  into  four  equal 
dihedral  t   angles.      For  convenience  of  reference  the    upper  and   lower 

*  Generally  two  ;  three  or  more  for  special  constructions. 
\/iti,  two,  eSpa,  hasf  or  /</<-<■ — angles  bounded  by  two  faces. 


2  ELEMi'lNTS  OF  bE^TPIPTIVE   GEOMETRY. 

division  of  the  vertical  plane  are  lettered  V  and  —  V  respectively,  the 
front  and  back  divisions  of  the  horizontal  plane  H  and  —  H.  With  a 
like  view  to  convenience  the  angles  are  numbered : 

I,  the  angle  above  //and  in  front  of  V\ 
II,  the  angle  above  H  and  behind  V\ 

III,  the  angle  below  H  and  behind  V; 

IV,  the  angle  be/oza  //"and  in  front  of  K 

THE   POINT   IN   SPACE. 

5.  The  planes  of  projections  being  thus  given,  the  projections,  so 
called,  of  any  point  in  space  are  determined  by  letting  fall  from  that 
point  perpendiculars  to  the  two  planes. 

Thus  from  a  (Fig.  2)  let  fall  the  perpendicular  aa" — termed   the  pro- 


a 

Fig.  2^ 
^_CC_ 

i 

-i 

^y^Y — 

.ci^ 

a 

1 

J 

Jecting  line — to  the  horizontal  plane,  H\  the  foot  a"  of  this  line  is  the 
horizontal  projection  of   the  point. 

The  combined  horizontal  projections  of  the  different  points  of  any 
object  constitute  the  horizontal  projection  of  that  object. 

In  like  manner,  a  perpendicular,  aa' — the  projecting  line — drawn  from 
the  point  a  to  the  vertical  plane,  V,  marks  with  its  foot,  a\  the  vertical 
projection  of  that  point. 

The  combined  vertical  projections  of  the  different  points  of  any  ob- 
ject constitute  the  vertical  projection  of  that  object. 

It  is  evident  that  the  points  a'  and  a"  thus  determined  are  the  pro- 
jections of  all  points  lying  in  the  two  projecting  lines  aa'  and  aa" . 

6.  The  projecting  lines  aa' ,  aa"  are  respectively  perpendicular  to  V 
and  H\  hence  any  plane  which  passes  through  them  must  likewise  be 
perpendicular  to  those  planes,  and,  by  Geometry,  to  their  line  of  inter- 
section or  ground-line,  GL,  which  it  cuts  in  the  point  a.  The  angle 
a'aa",  which  measures  the  dihedral  angle  of  the  two  planes  of  projec- 
tion, is  thus  a  right  angle. 

7.  The  angles  aa'a  and  aa"a  being  by  construction  right  angles,  the 
iigure  aa'aa"  is  a  rectangle  ;    whence  it  follows  that — 


GENERAL    CONSIDERATIONS.  3 

(i)  The  distance  aa"  of  the  point  a  from  //  is  equal  to  a'a^  or  the 
distance  of  the  vertical  projection  a'  above  the  ground-line,  for  the  first 
and  second  angles,  and  below  for  the  third  and  fourth. 

(2)  The  distance  aa'  of  the  point  a  from  V  is  equal  to  a"a,  or  the 
distance  of  the  horizontal  projection  a"  in  front  of  the  ground-line  for 
the  first  and  fourth  angles,  and  behind  for  the  second  and  third. 

(3)  The  perpendiculars  to  GL,  drawn  through  the  projections  a'  and 
a",  intersect  it  in  a  common  point,  a. 

The  lines  a'a  and  a"(x  are  termed  the  corresponding  ordinaies  of  the 
point  ay  and  a  is  its  ground-point. 

8.  With  this  orthogonal*  system  of  projection  it  is  found  that — 

(i)  All  distances  from  the  vertical  plane  arc  measured  in  their  true 
lengths  upon  the  horizontal  plane. 

(2)  All  distances  from  the  horizontal  plane  are  measured  in  their  true 
lengths  upon  the  vertical  plane. 

(3)  By  means  of  the  combined  projections  the  position  of  the  point 
in  space  may  be  accurately  determined. 

9.  Heretofore,  the  planes  of  projection,  together  with  the  construc- 
tions, have  been  considered  solely  with  reference  to  their  relative  posi- 
tions in  space.  But,  in  order  to  represent  both  projections  of  an  object 
on  one  plane,  the  draici/ig  plane — in  other  words,  in  order  to  render  the 


/ 

^^      a' 

ei 

^ 

i 

L 

'            y^ 

w 

3^. 

^-^ 

^^ 

^^     V 

y^v'          ~1 

G 

1 

1           / 

y 

a 

drawings,  thus  made  upon  the  planes  at  right  angles  to  each  other,  avail 
able  for  practical  purposes — some  modification  must  be  made  in  the  posi- 
tion of  these  planes. 

10.  This  is  effected  by  revolving  either  plane  of  projection  around 
the  ground-line,  GL,  until  it  coincides  with  the  other  plane.  Thus  the 
plane  V  (Fig,  3)  is  rotated  backivards  in  the  direction  of  the  arrows, 
until  it  coincides  with  the  plane  //.  when  the  upper  portion.  J'  rests 
upon    the  back    portion,  —  H,  of    the    horizontal    plane,   and    the    lower 


*  o(i^6<i,  straight  or    right,  yavia,    aiiglv — rcferrini;    tu    the  rectangular    positiun    of    the    pro- 
jecting lines. 


4  ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 

portion,   —  V,   is    covered    by    the    front    portion,   H,    of    the    horizontal 
plane.* 

11.  As  the  result  of  this  rotation  all  that  portion  of  the  drawing  plane 
which  lies  above  the  ground-line  represents  the  negative  horizontal,  —  H, 
and  the  superimposed  vertical  plane,'  V\  while  all  below  the  ground-line 
is  the  negative  vertical,  —  V,  and  the  superimposed  horizontal  plane,  H. 

12.  When  the  vertical  plane  is  rotated  backwards  into  the  horizontal 
plane,  the  vertical  projection  a' ,  and  its  ordinate,  a' a,  are  carried  around 
with  it,  and  maintain  unaltered  their  position  to  GL.  Hence,  the  ordi- 
nates  a' a  and  a" a,  being  now  in  the  same  plane,  passing  through  the 
same  point,  a,  and  remaining  perpendicular  to  the  ground-line,  are  the 
prolongations  of  each  other.  Whence  it  follows  that  the  two  correspond- 
ing projections  of  any  point  in  space  will  always  lie  in  the  same  line  perpen- 
dicular to  the  ground-line.  Two  points,  one  in  the  vertical  and  the  other 
in  the  horizontal  plane,  cannot,  therefore,  be  assumed  at  will  to  repre- 
sent the  projections  of  a  point  in  space  unless  they  fall  in  the  same  per- 
pendicular to   GL. 

13.  Conversely,  the  two  projections  {a! ,  a")  fix  the  position  of  the  point 
in  space.  For  the  line  GL,  being  perpendicular  to  the  two  ordinates  a' a, 
a" a,  will  likewise  be  so  to  the  plane  passing  through  them ;  hence,  the 
planes  V  and  H  are  by  Geometry  perpendicular  to  this  same  plane ;  or, 
reciprocally,  the  plane  aa'aa"  is  perpendicular  to  V  and  H.  If,  then,  the 
planes  of  projection  be  restored  to  their  original  position,  and  through 
the  projections  a'  and  a"  perpendiculars  be  erected  to  them,  these  per- 
pendiculars will  lie  in  the  plane  of  aa'aa",  and,  of  necessity,  intei-sect  in 
a  point  of  which  the  projections  will  be  {a',  a"). 

14.  When,  from  the  two  projections,  it  is  thus  sought  to  determine  the 
nature  and  position  of  the  object  in  space,  the  planes  of  projection  must 
always  be  imagined  as  restored  to  their  rectangular  position;  it  is  only 
for  the  purpose  of  making  the  drawings  or  projections  from  the  given 
data  that  the  planes  are  made  to  coincide. 


*  Long  experience  has  shown  the  utility  of  requiring  students  to  analyze  the  earlier  problems  by 
jneans  of  cards  cut  to  represent  the  planes  of  projection  in  space. 


GENERAL    CON  SID  ERA  TIONS. 


CHANGE     OF     G  K  O  U  N  D  -  1. 1  X  E  . 


15.  It  is  frequently  desirable,  nay,  even  necessary  in  practice,  to  em- 
1)1()V  more  than  one  vertical  plane.  Thus,  GL,  6",/,,  and  GJ.^  (^ij<-  4)  may 
represent  the  j^round-lines  of  three  distinct  planes,  all  of  which  are  per- 
l)cndicular  to  the  horizontal  plane  while  assuming  any  position  to  one 
another. 

If,  then,  a"  and  b"  are  the  horizontal  projections  of  two  points  in  space, 
it  is  manifest  that,  by  the  application  of  Art.  (12),  a  and  b\  </,'  and  b', 
a.,'  and  b^'  are  the  vertical  projections  of  these  same  points  on  the  three 
separate  planes ;  a'a,  «/«i.  «t^s  ^"^  '^  A  ^i7^i.  ^a  /^t»  measuring  in  each  case 
the  distances  of  the  points  above  the  common  horizontal  plane. 

16.  Notation  of  the  Point. — Throughout  this  work  the  point  in  s|)ace 
will  be  designated  by  a  small  letter,  and  its  projections  by  the  same  letter 
accented.  Thus  the  point  a  in  space  is  one  whose  projections  are  a'  and 
a"  for  the   vertical  and  horizontal  planes  respectively. 

On  the  plus  planes  the  projections  will  be  indicated  bv  a  cross,  and 
on  the  minus  planes  by  a  small  circle.  Where  the  two  projections  coin- 
cide, which  will  take  place  in  the  second  and  fourth  angles  when  the  point 
lies  in  tiie  bisecting  plane,  the  cross   \\\\\  be  enclosed  within  the  circle. 

POSITIONS   OF   THE    POINT   IN    SPACE. 

17.  A  point,  in  revolving  around  the  ground-line,  may  assume  the  fol- 
lowing general   positions: 

(I)  On  //,  in  front  of   V  (Fig.  5). 


(2)  In  the  first 
is  nearer  H  than 

(3)  In  the  bis 

(4)  In    the    tl 
nearer  Tthan  // 

(5)  In  ['abo 
(6^  In    the 

case  it  is  nearei 


the  biseeting  plane,  in  which  case  it 

CISC  it   is  equidistant  from  ['and  //. 
isccting  i)lane,   in    which  case    it    is 


?/'<':•(    the    bisecting    plane,  in   which 


6  ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 

(7)  In  the  bisecting-  plane,  equidistant  from  both. 

(8)  In  the  second  dihedral  angle,  below  the  bisecting  plane,  when  it  is 
nearer  —  H  than  V. 

(9)  In  -  //  behind  V. 

(10)  In  the  third  dihedral  angle,  above  the  bisecting  plane,  when  it  is 
nearer  —  H  than  V. 

(11)  In  the  bisecting  plane,  equidistant  from  both. 

(12)  In  the  third  dihedral  angle,  below  the  bisecting  plane,  when  it  is 
nearer  —  V  than  —  H. 

(13)  In  -  F  below  ^. 

(14)  In  the  fourth  dihedral  angle,  belozv  the  bisecting  plane,  when  it  is 
nearer  —  V  than  H. 

(15)  In  the  bisecting  plane,  equidistant  from  both. 

(16)  In  the  fourth  dihedral  angle,  above  the  bisecting  plane,  when  it  is 
nearer  H  than  —  V. 

(17)  In  the  ground-line,  when  it  is  a  point  of  both  planes. 

18.  Problem. —  To  find  the  projections  of  a  point,  its  distances  from  the 
planes  of  projection  being  given. 

Let  the  point  (Fig.  6)  be  assumed  to  be  half  an  inch  from  H  and  one 
inch  from  V. 

V  having  been  revolved  (Art.    10),  draw  an  indefinite  horizontal  line, 


xa 


the  ground-line  GL.  Since  the  two  projections  must  always  lie  in  the 
same  perpendicular  to  GL  (Art.  12),  draw  an  indefinite  ordinate  in  that 
position,  and  lay  off  the  distances  half  an  inch  above  and  one  inch  be- 
low GL ;  the  points  («',  a")  thus  determined  are  the  projections  sought. 
(Art.  8.) 

19.  Analyze,  from  the  given  projections,  the  positions  of  the  points  in  space 

(Fig-  7). 

20.  Problem. — Fro^n  the  data  (Art.  18)  to  find  the  projections   of  a  point 
in  the  remaining  diJiedral  angles. 

21.  Probleinl — To  find  tJic  projrrf^'-^'-     '"  g  in  the   bisecting 
planes  of  the  first  and  foiirtJi  diJiedral 

22.  Problem. —  To  find  the  projt  in  either  plane  of 
projection. 


GENERAL    CONSIDER  A  TIONS. 


Kis..  7 


23.  Prorlem. —  The  primitive  projections  of  a  point  being  given,  to  find  its 
new  vertical  projection  upon  any  neiv  vertical  plane. 

(i)  By  Art.  (12)  the  horizontal  projection  and  the  new  vertical  pro- 
jection ought  to  lie  m  the  same  perpendicular  to  GL. 

(2)  The  distance  of  the  point  from  //"remains  unaltered.  Hence  if  {a\a") 
(Fig.  8)  are  the  primitive  projections  of  the  point  <?,  let  fall  upon  the  new 


ij^round-line  C.Z,  the  ordinate  a" 01^,  and  lay  off  a  distance  a^a'  equal  to  aa\ 
above  or  below  6^,Z,  as  the  point  a'  may  be  above -or  below  GL.  The 
point  rt,'  is  the  new  vertical  projection  sought. 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


CHAPTER  II. 


THE  LINE. 


24.  A  right  line  in  space  may  have  one  of  three  positions  to  the  planes 
of  projection  (Fig.  9) : 

(i)  Parallel  to  both,  i.e.,  their  line  of  intersection,  or  ground-line  {a). 

/  X   T3       11  1   4.  i  perpendicular  to  the  other  ib) ; 

(2)  Parallel  to  one,    {V    }.   .        ,      , ,         . ,        ,  \ 
^  ^  '     (  inchnmg  to  the  other  {c). 

i  inclining  at  any  angle  {d^ ; 

(3)  Parallel  to  neither,   -  in  a  plane  perpendicular  to  GL  (e) ; 

(  intersecting  GL  (/). 

25.  As  a  line  is  >i  succession  of  points,  its  projection  on  any  plane  will 
be  determined  by  projecting  each  point  of  the  line  on  that  plane.  Thus, 
if  from  the  different  points  of  the  line  AB  (Fig.  9,  a)  perpendiculars  or 
projectmg  lines  be  drawn  to  N,  their  feet  will  indicate  the  horizontal  pro- 


ng, o 


\ 


Qisff^i 


jections  of  those   points,   and   the   line  a"d",  which    passes   through  them, 
the  horizontal  projection  of  the  line  itself. 

In  like  manner  the  vertical  projection  a'b'  will  be  found  by  drawing 
projecting  lines  to  the  vertical  plane. 

26.  This  regular  succession  of  parallel  projecting  lines  will  form  a  sur- 
face, which  in  the  case  of  the  right  line  is  a  projecting  plane,  but  with 
the  curved  line   a  projecting  cylinder. 

In  either  case  the  projection  of  the  line  lies  in  the  lijie  of  intersection 
betiveen  the  projecting  surfaces  and  the  planes  of  projection. 

27.  Hence  the  projection  of  a  right  line  is  a  right  line, — with  the 
single  exception  (Fig.  9,  h\ — and  is  obtained  by  passing  through  the  line 
in  space  planes  respectively  perpendicular  to  the  coordinate  planes  of  pro- 
jection. • 


THE  LINE.  9 

28.  As  two  points  not  consecutive  geometrically  determine  the  posi- 
tion of  any  right  line,  so  the  projections  of  any  two  of  its  pcjints  de- 
termine its  projections. 

29.  In  general,  a  right  line  will  be  fully  determined  by  its  two  pro- 
jections; for  if  the  coordinate  planes  be  restored  to  their  rectangular 
}>osition,  and  through  each  projection  of  the  line  a  plane  be  erected  jicr- 
pendicular  to  the  coordinate  planes  respectively,  the  two  planes  thus 
constructed  will  each  contain  the  line  in  space  which  must  be,  conse- 
([uently,  their  line  of  intersection.  When  both  projections  are,  however, 
perpendicular  to  GL  (Fig.  9,^,/),  the  projecting  planes  are  also  perpen- 
dicular to  GL,  and  coincide;  hence  they  will  give  no  line  of  intersection, 
and  fail  to  fix  the  line  in  space. 


30.  Problem. —  The  projections  of  tii'o  points  being  given,  to  find  the  pro- 
jections of  the  line  which  they  determine. 

1st  Case. — When  the  projections  do  not  lie  in  the  same  perpendicular 
to  GL  (Fig.  10). 

Let  {a\  a")  and  {d',  b")  be  the  given  points.  The  projections  must  pass 
through  the  corresponding  projections  of  the  points  (Art.  28).  hence  a'b' 
and  a"b" ,  are  the  vertical  and  the  horizontal  projections  sought. 

3^'ig.ll  Kig.ia  Kig.  13 

«'k  fa-  aV6' 

i  !  i 


a^ 


"1^  a,fi 

I  I  i^ 

I  <  ! 


a"yl)' 


2d  Case. — Where  the  projections  of  the  points  lie  in  a  common  per- 
pendicular to  GL. 

If  the  projections  {a\a")  and  (^', //';  are  separate  points  (Fig.  11),  a'b' 
and  a"b"  are  again  the  required  projections  of  the  line,  but  the  project- 
ing planes  coincide  and  are  perpendicular  to  GL. 

If  a"  and  /^"coincide  (Fig.   12),  and  <^' and  b'  are  separate,  the   entire 


lO 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


line  is  projected  horizontally  in  a  pointy  and  vertically  in  a  line  a'b'  par- 
pendiciilar  to  GL. 

If  a'  and  //  coincide  (Fig.  13),  and  a"  and  //'  are  separate,  then  the 
entire  line  is  projected  vertically  in  a  point,  and  horizontally  in  a  line  per 
pendicular  to  GL. 

The  corresponding  projections  a' ,  b'  and  a" ,  b"  cannot  both  respec- 
tively coincide,  since  the  line  in  space  cannot  assume,  at  one  and  the  same 
time,  a  vertical  and  a  horizontal  position  to  the  coordinate  planes. 

31.  Problem. — Given  the  projections  of  two  points,  to  find  the  projections 
of  a  third  point  of  the  line  which  they  determine. 

\st  Case. — Both  projections  of  the  line  may  incline  to  GL.  Let  {a' ,  a") 
and  (//,  //')  be  the  given  points  (Fig.  14).  Draw  the  projections  of  the 
line  passing  through  them.  Any  point  c  of  the  "line  ab  in  space  gives  a 
horizontal  projection  somewhere  on  the  line  a"b",  and  a  vertical  projec- 
tion somewhere  on  the  line  a'b',  since  the  projections  of  the  line  contain 


Fig.  14. 


^a" 


all  the  projections  of  its  successive  points.  But  as  the  projections  of  a 
point  alwa3^s  lie  in  the  same  perpendicular  to  GL,  the  projections  of  the 
third  point  will  be  found  by  erecting  a  perpendicular  to  GL  and  mark- 
ing its  points  of  intersection  {c,  c")  with  {a'b',  a"b"). 

2d  Case. — The  line  ab  may  lie  in  a  plane  perpendicular  to  GL. 

As  in  the  preceding  case  the  point  c  will  give  projections  which  are 
points  of  the  projections  of  the  line  ab.  But  as  the  two  projections  He 
in  a  common  perpendicular  to  GL,  the  ordinate  drawn  from  any  point 
coincides  with  that  perpendicular,  and  hence  determines  no  point  of  inter- 
section.    (Fig.  15.) 

Assume,  then,  a  new  ground-line  G,L,  which  shall  not  be  parallel  to 
the  primitive  one,  and  find  the  new  vertical  projections  (^/,  ^/)  and  the 
projection  of  any  intermediate  point  r/  of  the  line  joining  them.  Set  off 
the  distance  y^c^'  thus  obtained  from  a  to  c'  and  {c' ,  c")  are  the  required 
projections. 

In  practice,  whenever  the  conditions  of  the  problem  render  it  prac- 
ticable, the  new  ground-line  ought  to  be  taken  at  right  angles  to  the 
primitive  one  (Fig.  16),  thereby  avoiding  more  or  less  complex  construc- 
tions. 

■x^d  Case. — The  line  may  be  perpendicular  to  either  'plane. 


THE   LINE. 


II 


In  either  case  the  solution  is  immediate.  Should  the  line  be  vertical 
(Fit^.  17),  c"  will  coincide  with  the  horizontal  projection  a"b'\  wherever 
the  projection  c'  may  be  assumed. 

Should  the  line  be  horizontal  (Fig.  i8j,  c'  will  coincide  with  the  ver- 
tical projection  a'b\  wherever  c"  may  be  assumed. 

32.  1  he  preceding  case  leads  to  the  following 


10    Y 


c' 


4(.a" 


a"%h' 
c" 


Pkohlkm. — A  line  being  given,  to  find  upon  that  line  ii)  the  vertical  pro- 
jection of  any  given  point ;  (2 )  the  horizontal  projection  of  any  given  point. 

(i)  The  distance  of  any  point  from  the  horizontal  plane  is  the  meas- 
ure of  the  distance  of  the  vertical  projection  from  GL\  hence,  to  deter- 
niine  upon  the  vertical  projection  of  the  line  that  point  whose  distance 
is  known,  draw  a  line  c'd'  (Fig.  19)  parallel  to  GL  and  at  the  recpiired 
distance  from  it;  the  point  of  intersection  {a\  a")  is  the  one  sought. 

(2)  A  similar  construction,  c"d",  determines  by  its  point  of  intersection 
{a' ,  a")  the  point  at  a  given  distance  from  V. 


33,  Problem. — A  line  being  given,  to  find  upon  that  line  (i)  the  point 
whose  distance  from  H  is  a  minimum ;  (2)  the  point  whose  distance  from  V 
is  a  minimum. 

(i)  From  the  preceding  case  it  is  evident  that  the  nearer  the  extremity 
of  the  projection  a'b'  approaches  GL,  the  nearer  it  is  to  H\  hence,  when 
the  projection  is  prolonged  until  it  intersects  GL,  the  point  ot  intersection 
//'  is  the  vertical  projection  of  the  required  point  (Fig.  21)  from  which  the 
horizontal  projection  //"  may  be  determined  by  means  of  the  ordinate. 

The  point  {li\  h")  whose   distance    from  //"  is   a   minimum   is   that  in 


12  ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 

which  the  line  ab  in  space  pierces  //,  and  is  termed  the  horizontal  trace  or 
piercing-point  of  that  hne. 

(2)  The  nearer  the  extremity  of  the  projection  a"b"  (Figs.  21,  22,  23} 
approaches  GL  the  nearer  is  the  line  in  space  to  V\  hence,  when  the 
projection  is  prolonged  until  it  intersects  GL,  the  point  of  intersection  v" 


--ill 


is  the  horizontal  projection  of  the  required  point  from  which  the  vertical 
projection  v'  may  be  determined  by  means  of  the  ordinate. 

The  point  (t/,  v"^  whose  distance  from  V  is  a  minimum  is  that  in 
which  the  line  ab  in  space  pierces  V,  and  is  termed  the  vertical  trace  or 
piercing-point  of  that  line. 

When  the  given  line  lies  in  a  plane  perpendicular  to  GL  (Figs.  22,  23)^ 


Fig.  22 


a 


Fig.  23 


ii 


determine  the  piercing-points  on  a  new  vertical   plane,  and   transfer  th( 

measurements  thus  obtained  upon  the  primitive  planes  of  projection. 

In  Fig.  23  the  new  vertical  plane  coincides  with  the  plane  of  the  Hne» 
With    geometrical    lines    capable    of    unlimited     extension    these    two 

piercing-points  are  pecuharly  fitted  to  fix  the  position  of  the  line  in  space 

to  the  planes  of  projection. 

34.  The  preceding  case  may  assume  the  following  form : 

Problem. —  To  find  the  traces  or  piercing-points  of  a  line  the  projections  of 

which  are  given. 


THE   LINE.  13 

35.  Applying  the  principle  of  picrcini^-points  to  the  cases  of  Art.  (24;. 
it  will  be  seen  that— 

When  the  line  in  space  intersects  neither  plane  of  projection  it  must  be 
parallel  to  both,  and  hence  to  their  line  of  intersection,  or  GL.  In  such 
a  ])()siti()n  both  projections  are  parallel  to  GL,  since  under  any  other  sup- 
jiosition  they  would  intersect  it  and  thus  make  the  line  in  space  pierce 
the  coordinate  planes,  which  is  contrary  to  the  conditions  of  the  problem. 

Nine  cases  are  to  be  noted  (Fig.  24)  ;  fotir  as  the  line  lies  in  either 
dihedral  angle,  four  as  it  lies  in  either  coordinate  plane,  and  otic  as  it  lies 
in  the  ground-line.  When  the  line  is  equidistant  from  the  coordinate 
planes  it  lies  in  the  bisecting  planes ;  hence  in  the  second  and  fourth 
angles  the  i>rojections  coincide. 

Kig.  24 


h' 


36.  [Graphical  Characteristics  of  Lines. — As  the  right  line  is  fixed  in 
position  by  anv  two  points  which  are  not  consecutive,  in  a  similar  way 
its  projections  will  be  indicated.  Thus  ab  will  express  a  line  in  space, 
and  a'b' ,  a"b"  its  vertical  and    horizontal  projections  respectively. 

It  is  customarv  to  distinguish  between  the  various  lines  which  are  the 
products  of  the  constructions,  as,  for  instance,  between  those  which  are 
seen  or  concealed,  and,  again,  between  those  which  indicate  the  given 
data  and  the  results,  and  those  which  are  simply  the  means  of  their  attain- 
ment.    Accordingly  they  may  be  classified  and  delineated  as — 

(i)  Principal  lines,  the  graphical  representations  of  the  data  and  the 
results  attained  therefrom,  drawn  /////  when  seen  and  clotted  when  hidden, 
whether  by  the  planes  of  projection,  portions  of  an  object  of  which  they 
form  a  ])art,  or  by  the  interposition  of  other  objects. 

(2)  Const  met  ion  lines,  employed  as  auxiliaries  in  determining  the  re- 
quired solution  or  connecting  corresponding  projections  of  the  several 
points,  lines,  etc.,  and  drawn  as  broken  lines  composed  of  short  dashes.] 

37.  When  the  line  in  space  intersects  but  one  of  the  coordinate  planes 
it  is,  of  necessity,  parallel  to  the  other,  whatever  the  position  it  assumes 
to  the  plane  it  intersects.  Should  it  be  a  vertical  line,  its  horizontal 
projection  is  reduced  to  a  point  (F\g.  9,  /'),  since  the  projecting  lines  and 
the  line  in  space  coincide.  The  vertical  projection  is  a  line  perpendicular 
to  the  ground-line,  s\\\CQ  the  vertical  projecting  plane  and  the  vertical  plane 
of  projection  are  both  perpendicular  to  H. 


14  ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 

There  are  three  general  positions  to  be  noted  for  this  case,  according- 
as  the  line  is  situated  in  front  of  F,  in  that  plane  or  behind  it.     (Fig.  25.) 

Three  similar  positions  (Fig.  26)  are  indicated  when  the  line  is  perpen- 
dicular to  V,  according  as  it  is  situated  above,  in  or  below  H. 


ITigr.  as 


a 

a' 

a' 

(/'oV 

b' 

I) 

&' 

a' 

h" 

a'xb 


Fig.  26 


a'|6' 


a 

6' 

Ij' 

h" 

h" 

a'ob' 

a 

a" 

a" 

38.  A  line  in  space  parallel  to  either  plane  of  projection  and  inclining  to  the 
other  is  projected  on  the  former  in  a  line  parallel  and  equal  to  itself,  a?id  on 
the  latter  in  a  line  parallel  to  GL. 

Should  the  line  be  parallel  to  F(Fig.  27,  i),  then  the  vertical  projection 
a'b'  will  be  parallel  to  the  line  ab  in  space,  since  these  two  lines,  lying 
in  the  same  projecting  plane,  cannot,  by  the  conditions  of  the  case,  inter- 
sect. They  are  equal  inasmuch  as  they  are  the  opposite  sides  of  the 
rectangle  aba'b' . 

The  horizontal  projection  a''b"  will  be  parallel  to  GL  inasmuch  as  the 


projecting  plane  and   V  are  parallel,  and  hence  are  cut  by  H  in  parallel 
lines. 

Should  the  line  ab  be  parallel  to  H  (Fig.  27,  2),  by  similar  reasoning 
it  may  be  shown  that  the  horizontal  projection  a"b"  will  be  parallel  and 
equal  to  ab,  and  the  vertical  projection  a'b'  parallel  to  GL. 

39.  Fig.  28  exhibits  the  various  general  positions  which  this  case  may 
assume  as  the  line  ab  in  space  is  located  above,  in  or  below  H,  in  front 
of,  in  or  behind   V. 

(i)  Above  H,  oblique  to   V',  (4)  In  front  of   F,  oblique  to  H\ 

(2)  In  H,  oblique  to   V\  (5)  In    F,  oblique  to  H\ 

(3)  Below  H,  oblique  to   V\  (6)  Behind  V,  oblique  to  H. 


TIfF.   I.IXE.  ii; 

40.  Any  line  lying  in  the  bisecting  planes  of  the  dihedral  angles  will 
be  projected  in  the  first  and  third  angles  in  lines  which  incline  in  the 
same  direction  and  at  the  same  angle  to  GL,  and  in  the  second  and  fourth 
in  lines  which  coincide. 

From  this  it  results  that  any  line  parallel  to  a  bisecting  plane — that 
is,  parallel  to  any  line  of  that  plane — will  be  projected  in  the  first  and 
third  angles  as  above,  and  in  the  second  and  fourth  in  parallel  lines. 


Kig.  S8 


a'        r, 


a" 


I)' 


41.  When  the  line  in  space  intersects  both  coordinate  planes  it  inclines 
to  both,  atui  gives  projections  luhich  are  shorter  than  itself  and  7chieh  incline 
to  GL. 

Thus,  in  Fig.  29  the  line  ab  is  the  hypothenuse  of  the  right-angled 
triangles  abe  and  abd,  hence  is  greater  than  the  bases  ac  and  bd  or  their 
equivalents,  the  projections  a"b"  and  a'b' . 

Again,  the  projecting  planes  of  the  line  ab  incline  to  GL,  cut  the  co- 
ordinate planes  in  inclining  lines,  and  hence  give  inclining  projections. 


Exceptions  arise  when  the  line  lies  in  a  profile  plane  perpendicular  to 
GL  (Fig.  9,  e,  /),  in  which  cases  both  projections  He  in  the  traces,  and 
consequently  are  perpendicular  to  GL. 

In  case  e  there  are  four  positions  for  the  line,  as  a  portion  ot  it  is  inter- 
cepted in  any  one  of  the  four  dihedral  angles. 

In  case  f,  should  the  line  lie  in  the  bisecting  plane,  its  projections  in 
the  second  and  the  fourth  angles  will   respectively  coincide. 

42.  Analyze  the  positions  of  the  lines  given  by  their  projections 
(Fig.  301 


i6 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


T^ig.  SO 


d^' 


43.  Problem. —  To  find  the  projections  of  a  right  line  parallel  to  GL,  i^" 
long,  f "  from  H  and  \"  from  V. 

(i)  Draw  lightly  an  indefinite  ordinate  perpendicular  to  GL  (Fig.  31). 

(2)  Lay  off  upon  this  line  a  distance  of  three  quarters  of  an  inch  above 
and  half  an  inch  below  GL,  thus  determining  the  projections  {a' ,  a"). 

(3)  Draw  two  full  lines,  a'b' ,  a"b" ,   \\"  in  length,  through  these  points 
and  parallel  to  GL ;  the  lines  so  drawn  are  the  projections  required. 

From  the  given  data  determine  the  projections  in  the  remaining  three 
angles. 


n: 

S' 

31 

I 

1 

G 

1 

L 

ja 

/3 

i 
i/i" 

44.  Problem. —  To  find  the  projections  of  a  right  line  i"  long,  ^'  from 
either  plane  and  perpendicular  to  it. 

Let  the  line  be  vertical  (Fig.   32,    i). 

(i)  Draw  an  indefinite  ordinate  perpendicular  to  GL. 

(2)  Mark  upon  this  line  two  points,  a  ,  b' ,  respectively  indicating  in 
their  distances  from  GL  the  heights  of  the  two  extremities  of  the  line 
above  H. 

(3)  On  this  same  ordinate  lay  off  a  distance  aa"  equal  to  the  distance 
of  the  line  from  V.     Then  will  {a'b' ,  a"b")  be  the  projections  required. 

From  the  given  data  determine  the  projections  in  the  other  dihedral 
angles. 

45.  Problem. —  To  find  the  projections  of  a  line  parallel  to  07ie  plane  of 
J>rojection,  and  inclining  to  the  other. 

Let  the  line  be  i^"  long,  parallel  to  V,  and  inclining  at  any  angle  a  to 
H\  i"  from  Fand  \"  from  H. 


THE   LINE.  17 

(i)  Draw  an  indefinite  ordinate  (Fig.  32,  2),  and  mark  the  projections 
a'  one  quarter  of  an  inch  above,  and  a"  one  inch  below,  GL,  thus  indicating 
the   projections  of  the  lower  extremity  of  the   line   in   si)ace. 

(2)  Draw  through  a'  the  vertical  pnjjection  a'b\  one  and  a  half  inches 
long  and   making  the  angle   ix   with    GL. 

(3)  Through  a"  draw  the  horizontal  projection  a"b"  parallel  to  GL,  and 
limited   by   the  ordinate   let   fall   from   b' . 

From  the  given  data  draw  the  projections  in  the  other  dihedral  angles. 


46.  Problem. —  To  find  the  projections  of  any  line,  its  piercing-points  being- 
given. 

Let  a'   and  //'   be   the  given   piercing-points  (Fig.  33). 

The  point  rt' being  in  /'  is  its  own  vertical  projection,  its  horizontal 
projection,  a",  falling  in  GL.  Similarly.  /;"  being  a  point  in  H  is  its  own 
horizontal  projection,  its  vertical  projection,  //,  falling  in  GL.     Joining  the 


Kig.  33 


corresponding  projections  a'  and  b\  a"  and  //',  {a'b\  a"b")  are  the  projec- 
tions required. 

Solve  the  remaining  three  cases  (Fig.  33). 

47.  Problem.—^  /ine  being  given  by  its  projections,  to  find  a  point  of 
that  line  ivhosc  distances  from  the  coordinate  planes  shall  be  in  a  given  ratio. 

Let  the  ratio  be  as  x  :  y,  and  let  (a'b\  a"b")  be  the  given  line  (Fig.  34). 

Draw    any    perpendicular    to    GL    intersecting    the    projections   in    the 


i8 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


points  c' ,  c" ,  and  divide  c'c"  in  such  a  way  that  the  two  parts  shall  bear 
to  each  other  the  given  ratio.  It  is  manifest  that  this  may  be  effected 
in  two  ways,  as  the  point  of  division,  a,  lies  between  or  beyond  the  points 
c',c".  The  proportions  then  will  be  c"a  :  c'a  ::  x  \  y  for  a  intermediate,  and 
a/'  :  oi^c'  '.',  X  '. }/  for  ofj.     Join  a  and  a,  with  d',  the  point  in  which  the  pro- 


jections cross,  and  draw  ordinates  through  the  points  e  and  e  where  these 
lines  cut  GL ;  the  points  {e' ,  e"),  in  which  these  ordinates  intersect  the 
given  projections,  are  the  projections  of  the  required  point. 

48.  Problem. — A  line  being  given  by  its  projections,  to  find  a  point  of  tJiat 
line  whose  distances  from   the  coordinate  planes  shall  be  eqjial. 

As  in  the  preceding  case,  draw  any  perpendicular  c'c"  to  GL  (Fig.  35). 


^     \ 


Mark  its  intersections  c' ,  c"  with  a'b' ,  a"b" ,  and  divide  c'c"  so  that  the 
parts  shall  be  in  the  given  ratio  of  1:1.  Join  the  division-point,  a,  with 
d\  and  erect  an  ordinate  at  the  point  where  the  line  thus  obtained  cuts 
GL\   the  points  {e' ,  e")  are  the  projections  required. 

In   this   case   the    second    division-point   is   at   infinity ;    hence  the   line 
joining  it  with  d'  is  a  perpendicular  to  GL,  thus  giving  {d',  d"). 


LINES,   PLANE   FIGURES  AND  PLANES. 


I^ 


CHAPTER  III. 
LINES,  PLANE  FIGURES  AND  PLANES. 

49.  Two  lines  in  space  may  assume  two  <^eneral  positions  to  each 
other:  they  may  either  lie  in  the  same  plane,  or  they  may  have  such  a 
position  that  no  plane  can  be  passed  through  them.  In  the  first  case  they 
may  either  intersect  or  be  parallel ;  m  the  second  case  they  can  affect 
neither  of  these  positions. 

PARALLEL    LINES. 

50.  Two  parallel  lines  in  spaee  give  projections  that  are  parallel,  since 
their  projecting  planes  are  parallel,  and  hence  cut  the  coordinate  planes 
in  parallel   lines. 

Should  either  pair  of  projections  be  reduced  to  points,  the  lines  in 
sj)ace  are  parallel,  since  they  are  perpendicular  to  the  same  plane. 

51.  Conversely,  zvhen  the  projections  of  the  same  name  are  parallel,  the 
lines  in  space  are  parallel,  unless  they  lie   in   planes  perpendicular  to  GLy 


L.         G 


Xn"    ^b 


in  which  case  the  true  position  can  be  determined  bv  a  change  of  ground- 
Imc. 

^.    Prohi.ENL — Through  a  given  point  to  pass  a  line  parallel  to  a  given  line. 

Three  cases  are  presented  for  solution  : 

(i)  Both  projections  of  the  line  may  incline  to  GL. 

Let  (/',  p")  be  the  given  point  (Fig.  36K 

Through  p'  and  /'  draw  parallels  respectively  to  a'b',  a"b",  the  projec- 
tions of  the  given  line. 

(2)  The  line  may  lie  in  a  plane  perjiendicular  to  GL  (Fig.  IJ^. 

Assume  a  new  ground-line.  6",/.,.  and  determine  the  new  vertical  pro- 


20 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


jections  //  and  a^b^  of  the  given  point  and  line.  Through  //  draw  p^ii' 
parallel  to  a^b^ ,  and  find  its  horizontal  projection  p"n"  parallel  to  a'^b" . 
Through  /'  draw  a  parallel  to  a'b' ,  and  determine  the  point  n'  by  laying 
off  a  distance  above  GL  equal  to  the  distance  of  w/  from   G^L^. 

(3)  The  line  may  be  perpendicular  to  either  coordinate  plane,  in  which 
case  the  solution  is  at  once  obtained. 

INTERSECTING   LINES. 

53.    Tzvo  lilies  in  space  which   intersect  give  projections   whicJi  intersect   in 
points  zvhich  lie   in   the  same  perpendicular  to  GL. 

For  the  point  of  intersection  (Fig.  38)  being  common  to  both  lines,  its 

33  ;Eig.  39 


projections  must  likewise  be  common  to  the  two  projections  and  follow 
the  law  which  governs  the  projections  of  any  point  (Art.  12). 

Conversely,  when  the  like  projections  of  the  lines  intersect  in  points 
lying  in  a  common  perpendicular  to  GL,  the  lines  in  space  intersect. 

Whence  it  follows,  that  intersecting  projections  cannot  be  assumed  at 
will  to  represent  the  projections  of  intersecting  lines  unless  the"  points  of 
intersection  are  in  conformity  with  the  above  condition  (Fig.  39). 


Fig.  4rO 


54.  Problem. — Analyze  the  positions  of  the  lines  in  space  whose  projections 
are  given  (Fig.  40). 

55.  Should  either  or  both  lines  lie  in  a  plane  perpendicular  to  GL,  a 
new  ground-line  will  be  required  to  determine  their  position  to  each 
other.     Whence  arises  the  following : 


LINES,  rLAXE   FIGURES  AXD   PLANES. 


21 


Problem. —  To  dctermme  the  position  of  tzuo  lines  ivhen  either  or  both  lie 
in  a  plane  perpendicular  to   GL. 

Let  {a'b',  a"b")  and  {c'd\  c"d")  be  the  given  lines  (Fig.  41).  Assume  a 
new  ground-line,  G",Z.,,  and  find  the  new  vertical  projections.  The  lines 
intersect,  inasmuch  as  //,  the  point  of  intersection  on  the  new  vertical 
plane,  and  /"  lie  in  a  common  perpendicular  to  GL. 


Fig.  42  exhibits  the  case  in  which  the  two  lines  ab  and  ed,  both  lying 
on  a  plane  perpendicular  to  GL,  are  found,  by  means  of  the  new  vertical 
projections,  to  be  parallel  to  each  other. 

56.  Problem. —  To  determine  the  projections  of  any  line  connecting  two 
given  lines  ivhich   intersect. 

Let  {a'b' ,  a"b")   and  {c'd',  c"d")  be   the  given   lines   (Fig.  43). 


,Ki-.  44 


Assume  anv  vertical  projection  ef'oi  the  required  line;  the  points  of 
uitersection  e"  and  / '  must  lie  in  perpendiculars  to  GL.  drawn  through  e' 
and  /'    respectivelv.      The    line    e'f"    is  the   horizontal   projection  sought. 

This   case    admits   of   an    infinite    number   of   solutions. 


22 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


57,  Problem. —  To  determine  the  relative  position  of  two  lines  zvliose 
projections  do  not  intersect  zuithin  the  limits  of  the  drawing. 

Let  {a'b\  a"b")  and  ic'd',  c" d")  be  the  two   lines  (Fig.  44). 

Assume,  as  in  the  preceding  case,  any  two  lines  a'd'  and  b'c'  intersect- 
ing the  given  lines,  and  find  their  .horizontal  projections  a" d"  and  b"c" . 
Should  the  given  lines  intersect  in  space,  the  assumed  lines  will  lie  in 
their  common  plane  and  also  intersect,  and  hence  their  points  of  inter- 
section (/',  p")  fall  in  the  same  perpendicular  to  GL. 

Should  the  lines  in  space  not  intersect  (Fig.  45),  the  points  in  which 
the  projections  cross  each  other  will  not  lie  in  the  same  perpendicular 
to  GL. 


58.  Problem. —  To  draw  the  projections  of  a  line  passing  through  a  given- 
point  and  intersectiftg  a  given  line,  when  either  projection  of  the  point  of 
intersection  lies  beyond  the  limits  of  the  drazving. 

Let  {a'b' ,  a"b")  be  the  given  line  (Fig.  46),  and  (/',  /")  the  given  point. 

Assume  p'a' ,  the  vertical  projection  of  the  line  required.  Construct  any 
triangle  p"m"n",  one  point  lying  m  the  horizontal  projection  a"b",  and 
the  other  in  the  ordinate  passing  through  the  pomt  of  intersection  a'. 
Construct  a  second  triangle  in  which  the  side  r"s"  is  parallel  to  m"n" , 
and  through  r"  and  s"  draw  parallels  respectively  to  m"p"  and  n"p'\ 
thus  determining  the  point  t" ,  The  line  joming  /''  and  t"  is  the  horizontal 
projection  sought. 


Fig.  4r 


LINES,  PLANE   FIGURES  AND  PLANES.  23 

PLANE    riGUKES. 

59.  When  the  intersecting-  lines  exceed  two  in  number  and  lie  in  the 
same  plane,  they  may  enclose  2i  plane  figure  whose  projections  can  be 
readily   determined   by   the   preceding  explanations. 

In  projecting  any  plane  figure,  whether  rectilinear  or  curvilinear,  the 
projecting-  planes  form  prismatic  or  cylindric  surfaces  (Figs.  47,  48)  whose 
intersections  with  either  plane  of  projection  mark  the  projections. 

60.  Problem.— T'c'  find  the  projections  of  any  plane  figure  parallel  to  the 
vertical  plane. 

Assume  the  regular  octagon  to  be  the  given  figure  (Fig.  49). 
Conceive    the    projecting    planes    of   the    different    sides;    they    form 
in   space  a  prism  the  base  of  which  is  the  given  object.     When  such  a  i 


Fds.  49 


imaginar)'  prism  is  cut  by  \\  the  section — that  is,  the  vertical  projection — 
is,  by  Geometry,  equal  to  that  base.  The  projecting  plane  to  H  and 
the  plane  of  the  object  coincide  and  are  parallel  to  V\  hence  the  hori- 
zontal  projection  is  a  right  line  parallel  to   GL. 

61.  Problem. —  To  find  t lie  projections  of  a  circle  parallel  to  the  horizontal 
plane. 

In  this  case  the  projecting  surface  becomes  a  C3-linder  (Fig.  50),  the  base 
of  which  is  the  circle  in  space.  When  this  surface  is  intersected  by  H, 
the  section — that  is,  the  horizontal  projection — is,  by  Geometry,  a  circle. 
The  projecting  plane  to  ['  and  the  plane  of  the  circle  coincide  and  are 
parallel  to  H\  hence  the  vertical  projection  is  a  right  line  parallel  to  GL. 
From  the  two  preceding  cases  it  necessarily  follows  that — 
.-///)'  plane  figure  parallel  to  one  of  the  coordinate  planes  gives  a  projection 
on  that  plane  parallel  and  equal  to  itself,  and  on  the  other  plane  a  right  line 
parallel  to  GL. 

62.  Proble.nl — To  determine  the  projections  of  a  rectangle  the  plane  of 
IV hie h  is  perpendicular  to   I'  and  inclines  to  H  at  any  given  angle. 

Let  a  be  the  given  angle,  and  let  the  longer  side  of  the  rectangle  be 
perpendicular  to  V  (Fig.  51),  thus  making  the  smaller  side  parallel  to  that 
plane. 

Since  the  projecting  surface  to  ['  and  the  plane  of  the  object  coincide, 
draw  the  vertical  projection  ah'  equal  \.o  the  smaller  side  and  making  the 
given  angle  a  with  GL.     Draw  the  ordinates,  and  lay  off  on  them  at  the 


24 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


required  distance  from  V  the  full  length  «' V  of  the  longer  side,  that  side 
being  by  the  conditions  of  the  problem  parallel  to  H.  Connect  the  ex- 
tremities of  these  lines  to  complete  the  horizontal  projection. 

63.  Problem. —  To  determine  the  projections  of  a  circle  the  plane  of  zvhich 
is  perpendicular  to  H  and  inclines  to   V  at  any  given  angle. 

Since  the  projecting  surface  to  H  and  the  plane  of  the  circle  coincide, 
draw  the  horizontal  projection  b"d"  a  right  line  equal  to  the  diameter  and 
making  the  given  angle  a  with  GL.  Assume  a  new  vertical  plane  parallel 
to  the  circle,  and  find  the  new  vertical  projection.  Determine  the  hori- 
zontal projections  b" ,  a",  d",  etc.,  of  points  of  the  circumference,  and  find 
the  vertical  projections  by  the  usual  method.  The  section  of  the  project- 
ing cylinder  is,  by  Geometry,  the  curve  of  an  ellipse. 

64.  Should   the  plane  of  the  object  be  perpendicular  to  GL,  both  pro- 


Fig.  sa. 


Fig.  5S 


jections  will  be  right  lines  perpendicular  also  to  that  line,   since  the  pro- 
jecting planes  and  the  plane  of  the  object  coincide  in  space. 


PLANES. 

65.  A  plane  is  a  surface  which  is  generated  by  the  motion  of  a  right 
line  in  such  a  way  that  each  of  its  points  describes  in  space  a  line  parallel 
to  some  second  fixed  right  line.  The  moving  line  is  termed  the  genera- 
trix, the  fixed  one  the  directrix. 

66.  The  position  of  a  plane  in  space  may  be  determined  by  two  lines 
either  parallel  or  intersecting,  by  a  line  and  a  point  not  in  that  line,  or 
by  three  points  not  in  the  same  right  line. 

67.  As  planes  are  of  indefinite  extension,  they  must  intersect  one  or 
both  planes  of  projection.  The  lines  in  which  they  so  intersect  are  termed 
the  traces,  and  in  ordinary  practice  are  the  means  whereby  the  planes  are 
determined  in  position. 

68.  The  plane  in  space  may  assume  the  following  general  positions 
to  the  coordinate  planes :    it  may — 


LINES,    PLANE   FIGURES  AND  PLANES. 


25 


(i)  Pass  through  the  ground-line,  giving  no  traces,  and  hence  indeter- 
minable except  by  the  use  of  a  new  vertical  plane  (Fig.  53)  ; 

(2)  Be  parallel  to  one,  giving  but  one  trace  parallel  to  GL  on  that 
plane  to  which  it  is  not  parallel  (Fig.  54) ; 


KiK.  R3 


KiK. 


(3)  Be  parallel  to  GL  and  incline  to  both  planes,  when  both  traces  will 
be  parallel  to  that  line  (Fig.  55); 

(4)  Be  perpendicular  to  GL,  when  both  traces  will  be  perpendicular  to 
that  line — xSxo.  profile  plane  (Fig.  56); 


■yxs.  GO 


B^ie,57 


Fie.  C8 


(5)  Be  perpendicular  to  one  plane  and  incline  to  the  other,  when  one 
trace  will  be  perpendicular  to  GL  and  one  inclined  to  it  (Fig.  57); 

(6)  Incline  to  both  planes,  other  than  in  the  first  and  third  cases,  when 
both  traces  will  incline  to  GL  (Fig.  58). 

The  graphical  representations  of  these  cases  are  shown  in  Fig.  59. 


Fig,  so 


69.  The  traces  being  lines  in  the  coordinate  planes  are  their  own  pro 
jections  on  the  planes  in  which  they  are  contained,  their  other  projections 
being  in  GL. 


26 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


70.  Besides  the  traces  the  following  are  characteristic  lines  of  the 
planes  in  space : 

(i)  The  horizontals,  or  the  lines  parallel  to  //  (Fig.  60). 

(2)  The  verticals,  or  the  lines  parallel  to  V  (Fig.  61). 

(3)  The  lilies  of  greatest  declivity,  which  measure  the  angle  of  the  plane 
in  space  with  either  plane  of  projection  (Fig.  62). 

71.  With  reference  to  these  lines  it  is  to  be  observed  that — 

(i)  The  horizontals  are  necessarily  parallel  to  the  horizontal  trace ; 
otherwise  they  would  intersect  it  and,  hence,  the  horizontal  plane. 

(2)  The  verticals  are  necessarily  parallel  to  the  vertical  trace  for  a 
similar  reason. 

(3)  The  lines  of  greatest  declivity  are  perpendicular  to  the  respective 
traces  (Fig.  62). 

Thus,  the  line  po  drawn  perpendicular  to  the  horizontal  trace  HN 
measures  the  greatest  declivity  of  the  given  plane  with  H.  For,  if  from 
/  we    draw  pr,    inclining   to  the   horizontal   trace,  and  //"  perpendicular 


to  H\  then  pr  ">  po,  and  p"r  '^  p"o,  while  pp"  remains  unchanged.  But 
in  two  right-angled  triangles  having  the  same  perpendicular,  the  greater 
the  base,  the  smaller  the  angle  at  the  base. 

72.  The  line  //"  being  perpendicular  to  H,  the  triangle  pp"o  lies  in  a 
plane  perpendicular  both  to  H  and  to  the  horizontal  trace  HN.  Hence 
the  angle  of  greatest  declivity,  pop" ,  Avill  be  determined  by  passing  a 
plane  at  right  angles  to  the  trace  on  that  plane  to  which  the  inclination 
is  sought. 

Whence  it  follows  that  tJie  traces  do  not  necessarily  measure  the  angles  of 
inclination. 

73.  The  traces  ot  a  plane  serve  to  indicate  its  position  in  all  cases 
except  that  in  which  they  coincide  with  GL  (Fig.  53),  since  they  represent 
the  projections  of  two  right  lines  of  that  plane  which  must  either  intersect 
or  be  parallel  to  each  other. 

74.  When  the  traces  of  a  plane  are  not  parallel  they  must  intersect 
each  other  in  a  point  in  GL,  since  the  coordinate  planes  and  the  given 
plane  form  a  solid  angle  whose  vertex  is  the  point  of  intersection  or  the 
piercmg-point  of   GL  on  that  plane. 


LINES,    PLANE   FIGURES  AND   PLANES. 


?^i 


75.  Notation  of  the  Plane. — Planes  are  usually  represented  by  capital 
letters,  and  their  traces  by  the  prefixes  H  and  V.  Thus  (Fi^s.  53-58), 
P  denotes  the  plane  in  space,  and  HP  and  VP  the  horizontal  and  vertical 
traces  respectively. 

When  a  plane  is  j^iven  by  two  rij^ht  lines  it  is  indicated  by  the  letters 
of  the  lines;  thus,  plane  {ab,  cd),  which  signifies  the  plane  determined  by 
the  right  lines  ad  and  cd.  In  a  similar  way,  the  plane  {ab,  c)  indicates  a 
plane  determined  by  the  right  line  ab  and  point  e.  Lastly,  the  plane 
{a,  b,  c)  expresses  one  whose  position  is  fixed  by  the  three  points  in 
question. 

Visible  traces  are  drawn  full,  concealed  traces  by  a  short  dash  and 
two  dots  alternately. 

Auxiliary  planes  are  represented  by  traces  drawn  with  a  short  dash  and 
dot  alternately. 

76.  Problem. —  To  analyze  the  position  of  a  plane  given  by  its  traces 
(Fig.  63). 


77.  Problem. —  To  pass  through  a  line  in  space  a  plane  perpendicular  to 
either  coordinate  plane. 

If  perpendicular  to  H  (Fig.  64),  the  required  plane  will  coincide  with 
the   projecting  plane  of    the    line    to  H ;    hence    its    horziontal  trace.    HN, 


Kig.  64 


iTig.  00 


JFA' 


H.V 


my     I 
!  L 

i 


will  contain  the  horizontal  projection  of  the  line,  and  its  vertical  trace, 
riV,  will  be  jierpendicular  to  GL. 

If  perpendicular  to  V  (Fig.  65),  then,  for  similar  reasons,  tlie  vertical 
trace  will  contain  the  vertical  projection  of  the  line,  and  the  horizontal 
trace  will  be  perpendicular  to  GL. 

78.  Problenl — Given  three  points  not  in  the  same  right  line,  to  find 
any  line  of  the  plane  which  they  determine. 

Applying   a    principle    of    Geometry   that    "  any  line    lying    in    a    plane 


28 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


intersects  all  other  lines  of  that  plane  except  those  to  which  it  is  parallel,"^ 
the  following  cases  may  be  distinguished  : 

(i)  The  plane  («,  b,  c)  in  which  the  projections  of  the  three  points  do  not 
coincide  (Fig.  66).  Join  the  points  by  three  lines  {a'b',  a"b"),  {b'c',  b"c"y 
and  {c'a' ,  c"a"),  and  take  any  intermediate  points,  as  (<?',  o")  and  {p\  p"), 
on  the  lines  so  determined ;  the  line  connecting  these  points  is  the  line 
required. 

Again,  if  through  any  point  as  {o' ,  o")  of  the  line  ab  (Fig.  67)  a  parallel 


be  drawn  to  either  of  the  other  two  lines,  a  line  of  the  plane  will  have 
been  found. 

(2)  In  Fig.  68  the  two  lines  {b'c',b"c")  and  {c'a',  c"a")  alone  admit' of 
the  immediate  use  of  an  intermediate  point ;  the  solution,  however,  may 
be  effected  as  in  the  preceding  cases. 

Should   the   horizontal   projections   a"  and   b"   coincide   (Fig.    69),   the 


line  joming  them  is  vertical  in  position  ;  hence  the  plane  passing  through; 
the  given  points  is  likewise  vertical,  and  every  line  lying  in  that  plane 
is  projected  on  H  in  its  horizontal  trace.  From  this  it  follows  that  any 
two  points  as  {o",p")  determine  a  line  of  the  plane. 

Should   the  two  projections  a'  and  b'  coincide  (Fig.  70),  the  line  joining 


LINES,    PLANE  FIGURES  AND   PLANES. 


29 


them  is  perpendicular  to  V\  hence  the  plane  passing  through  the  given 
points  is  likewise  perpendicular  to  F,  and  every  line  lying  in  that  plane 
is  projected  on  V  in  its  vertical  trace.  From  this  it  follows  that  any  two 
points  {p' ,  p')  determine  the  vertical  projections  of  any  line  of  that  plane. 

(3)  Should  the  three  points  given  lie  in  a  profile  plane,  the  plane 
being  perpendicular  to  both  coordinate  planes,  every  line  of  that  plane 
is  projected  upon  the  traces  thereof  (Fig.  71).  Hence,  any  two  jjoints 
{o\  o")  and  {p',p")  assumed  in  these  traces  determine  a  line  of  the  plane. 

79.  Problem. — Given  the  projections  of  three  points  not  in  the  same  right 
line,  to  find  any  point  of  the  plane  which  they  determine. 

Draw  any  line  of  the  plane,  as  in  preceding  cases,  and  find  anv  p(jint 
of  that  line  by  the  usual  methods. 

80.  Problem. — Given  one  projection  of  a  line  lying  in  a  given  plane,  to 
find  the  other  projection. 

Let  P  be    the   given    plane,  and  h"v"  the    given    projection  (Figs.  '/2, 


73).  The  piercing-points  of  the  line  must  lie  in  the  traces  of  the  plane 
and  be  common  to  the  projections  of  the  line ;  hence,  the  horizontal 
piercing-point  //"  falls  at  the  intersections  of  HP  and  h"v".  The  ver- 
tical piercing-point,  being  a  point  of  the  vertical  plane  and  of  the  vertical 
trace    VP,  is  projected    horizontally  at  v"\   the  ordinate  drawn    from   this 


Kisi.r4. 


point  intersects  VP  in  the  point  v' ,  the  vertical  piercing-point  of  the 
given  line.  As  anv  point  in  H  is  vertically  projected  in  GL,  h'v'  is  the 
required  vertical  projection. 


30 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


8^.  Problem. — Solve  the  cases  of  Fig.  74  under  Art.  80. 

82.  Problem. — Given  the  traces  of  a  plane,  to  find  the  projections  (i)  of 
any   horizontal,  (2)  of  any  vertical. 

(i)  Let   VP  and  HP  be  the  given  traces  (Fig.  75). 

Since  the  horizontal  is  a  line  parallel  to  the  horizontal  trace  (Art.  71), 
and  hence  to  H,  its  horizontal  projection  v"h"  is  projected  parallel  to 
HP,  and  its  vertical  projection  parallel  to  GL.  Assume,  then,  any  hori- 
zontal projection,  determine  its  vertical  piercing-point  {v',  v"\  and  draw 
v'h'  parallel  to  GL. 

Fig.  re 


(2)  In  like  manner  to  determine  the  vertical,  draw  any  line  h'v^  (Fig. 
76)  parallel  to  VP,  find  {li ,  h")  the  horizontal  piercing-point,  and  draw 
k"v"  parallel  to  GL. 

83.  Problem. —  The  plane  beih£  given  by  three  points  not  itt  the  same 
right  line,  to  find  a  vertical  or  a  horizontal. 

Draw  the  horizontal  or  vertical  projection  parallel  to  GL  and  proceed 
as  in  Art.  78. 

84.  Problem.  —  Given  one  trace  of  a  plane  and  the  projections  of  any 
point  in  that  plane,  to  find  the  other  trace. 

Let  VP  be  the  given  trace  (Fig.  ']'j\  and  {a! ,  a")  the  given  point. 
Through  the  point  draw  the  vertit:al  iji'v' ,  h"v"),  and  find  its  piercing- 
point  h" ;   the  required  trace  HP  passes  through  it. 


Kig 


In  Fig.  78  the  vertical  trace  VP  has  been  determined  by  means  of 
the  horizontal  {a"v" ,  a'v'). 

85.  Problem. —  To  find  the  traces  of  a  plane  passing  through  two  inter- 
sectinor  Hues. 


LINES,    PLANE   FIGURES  AND   PLANES. 


JI 


Let  {a'^' ■>  <^"b")  and  (c'd\  c"d")  be  the  two  lines  intersecting  in  {p\  o") 
(Fig.  79).  As  the  piercing-points  of  hnes  are  always  in  the  traces  of 
the  plane  which  contains  them,  find  the  piercing-jx^ints^  and  pass  the 
trace's  VP  and  HP  through  their  vertical  and  horizontal  projections  re- 
si)ectively.  When  pn^longed  they  intersect  in  GL,  and  thus  veriiy  the 
construction. 

86.  Problem. —  To  pass  a  plane  through  three  given  points  not  in  the 
same  right  line. 

Let  a,  b,  e  be  the  given  points  in   space  (Fig.  80). 

Join  them  two  and  two  by  lines,  and  find  the  piercing-points  thereof; 
the  traces  of  the  required   plane    VP  and  HP  pass  through  them. 


VF 


If  the  construction  has  been  accurately  determined,  the  three  piercing- 
points  of  the  same  name  will  lie  in  the  respective  traces,  and  the  traces 
themselves  intersect  in  GL.  A  special  case  arises  when  the  line  joining 
two  oi  the  points  is  parallel  to  either  or  both   planes  of  projection. 

Should  the  line  be  parallel  to  V  or  //.  then  the  vertical  or  the  hori- 
zontal trace  of  the  required  plane  will  be  parallel  t(^  the  vertical  or 
horizontal   projection   of  that  line. 

Should  the  line  be  parallel  to  GL,  the  traces  will  also  be  parallel  to 
that  line. 

87.  Prohi.KM.— T^i?  pass  a  plane  through  a  given  point  and  line. 

A  line  passed  through  the  given  point  parallel  to  or  intersecting  the 
given  line  lies  in  the  required  plane.  The  piercing-points  of  the  auxiliary 
and  given  lines  determine  the  traces  of  the  plane. 

ciian(;k   of  grol'nd-i.ine. 

88.  Problem. — Given  a  plane  by  its  traees,  to  determine  a  nezc  vertieal 
trace  by  a  change  of  ground-line. 

Let  {VP,  HP)  be  the  given  traces,  and  GJ..  the  new  ground-line 
(Fig.  81). 

The  new  vertical  plane  intersects  the  traces  of  the  given  plane  in 
two  points,  h"  and  a  point  whose  horizontal  projection  is  v" .  Lay  of! 
the  distance  v"7\'  equal  t(^  7>":'\  and  //'  t','  is  the  new  vertical  trace  re- 
quired. 


32 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


Fig.  82  illustrates  the  case  in  which  the  new  vertical  plane  has  been 
taken  at  right  angles  to  the  primitive  vertical,  and   Fig.  83  the  case  in 


li^ig.  82 


Fig.  83 


which  the  new  vertical  plane  has  been  taken  parallel  to  the  primitive 
vertical,  the  new  vertical  trace  being  for  this  position  parallel  to  the 
given  vertical  trace. 


PARALLEL   AND   INTERSECTING  ^LANES. 


33 


CHAPTER    IV 


I.     PARALLEL  AND   INTERSECTING   PLANES. 

89.  Two  planes  may  assume  two  general  p<jsitions  to  each  other : 
thty  may  be  parallel  or  they  may  intersect. 

With  the  parallel  system,  whatever  the  positi<jn  to  the  coordinate 
planes,  the  traces  of  the  same  name  are  always  parallel. 

This  is  in  accordance  with  a  theorem  of  Geometry  which  proves 
that  when  two  parallel  planes  are  cut  by  a  third  plane  the  lines  of  in- 
tersection, i.e.,   the   traces,  are  parallel. 

90.  The  converse  of  this  proposition  is  also  true ;  /.  e.,  zchen  the  traces  of 
the  same  natne  are  parallel,  the  planes  in  space  are  parallel. 

An  exception  arises  in  the  case  of  traces  parallel  to  GL. 

91.  With  the  intersecting  system  of  planes,  the  lines  of  intersection 
may  assume  every  conceivable  position  as  the  planes  themselves  alter 
their  position  to  the  coordinate  planes. 

92.  Prohlem. —  To  pass  through  a  given  point  a  plane  parallel  to  a  given 
plane. 

Let  {VP,  HP)  be  the  given  plane,  and  {p',p")  the  given  point  (Fig.  84). 


Fig.  8-4: 


Pass  through  the  given  point  any  line  parallel  to  the  given  plane. 
The  line  {p'v" ,  p'v')  parallel  to  a  horizontal  of  that  plane  is  such  a  line.  Find 
the  vertical  piercing-point  {v',  v"),  and  draw  the  traces  T'C^,  HO  parallel  to 
VP,  HP,  respectively. 

93.  Problem. —  Through  a  given  line  to  pass  a  plane  parallel  to  a  second 
line. 

Let  {a'b',  a"b")  and  {c'd',  c"ii")  be  the  given  lines  (Fig.  85). 

Through  any  point  {p\  p")  of  the  first  line  draw  a  parallel  to  the 
second  line.  The  required  plane  contains  these  two  lines,  and  hence  the 
determination  of  their  piercing-points  fixes  the  traces  rP,  HP. 


34 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


94.  Problem. —  ThroWgh  a  given  point  to  pass  a  plane  parallel  to  two 
given  lines. 

Through  the  given  point  pass  parallels  to  the  given  lines,  hnd  their 
piercing-points,  and  draw  the  traces  which  pass  through  them. 


INTERSECTING    PLANES. 
95.    The  intersection  of   any  two   surfaces   is    determined,   in    gene 


rai,  * 


by  the  aid  of  auxiliary  secant  planes,  which  pass  through  the  surfaces 
and  cut  lines  upon  them.  The  points  common  to  the  lines  thus  cut 
are  common  to  both  surfaces  and,  hence,  to  their  line  of  intersection. 

When  the  intersecting  surfaces  are  planes,  the  intersection  is  a  right 
line,  for  the  determination  of  which  two  auxiliary  secant  planes  will 
ordinarily  prove  sufficient. 

96.  Problem.— 7b  find  the  line  of  intersection  between  tzvo  planes,  given 
by  their  traces. 


■Fig.  sr 


Kig.  86 


Let  (FP,  HP)  and  {VM,  HM)  be  the  given  traces  (Fig.  86). 

By  the  application  of  the  preceding  principles,  the  two  coordinate 
planes  may  be  considered  as  the  auxiliary  secant  planes.  Thus,  V  cuts 
the  two  given  planes  in  the  vertical  traces,  which  intersect  each  other 
in  (/',  /'),  while  H  cuts  them  in  the  traces  HP  and  HM,  which  inter- 
sect in  ip',  o").      But  the  points  thus  determined  are  the  piercing-points 


F-ig.  89 


Fig,  90 


of   the   line   of   intersection    sought   (Fig.   87),   the    projections   of    which 
may  be  found  by  Art.  46. 

Figs.  88-93  are   applications  to  special  cases. 

In  Fig.  88  the  second  plane  is  vertical  in  position. 

In  Fig.  89  it  is  perpendicular  to    V. 

In  Fig.  90  the  first  plane  is  vertical,  the  second  perpendicular  to   V. 


PARALLEL    AA'D   IXTERSECTING  PLANES. 


35 


In  Fig.  91   the  second  plane  is  perpendicular  to  GL. 

In  Fig.  92  the  line  of  intersectif>n  lies  in  a  plane  perpendicular  to  GL. 

In  Fig.  93  the  line  of  intersection  pierces  —  H. 


Fig.  01 


Kifr.  nc; 


In  Fig.  94  the  vertical  traces  are  parallel.  In  this  case  the  line 
of  intersection  gives  but  one  piercing-point  (/>",  p')  at  the  intersection 
of  the  horizontal  traces,  and  hence  is  a  vertical  of  each  of  the  given 
planes. 

Kig,  0-1 

^  '  Kig.DS 


The  vertical  projection  o'p'  is  therefore  parallel  to  the  vertical  traces, 
and  the  horizontal  projection  o"p"  is   parallel  to  GL  (Art.  82). 

In  Fig.  95  the  line  of  intersection  is  a  horizontal,  there  being  but 
one  piercing-point  {o  ,  o" )  on  V. 

97.  In  Figs.  96,  97,  the  traces  of  the  two  planes  intersect  in  a  com- 
mon point  {a' ,  a")  in   GL.     The   intersection   of  the  given    planes  by   the 

^^^'■««  Kig.07 

^^/  VAT 

m 


coordinate  planes  determines  this  point  alone  ;  hence,  an  additional  aux- 
iliarv  plane  must  be  employed  to  obtain  a  second  point  of  the  line  ol 
intersection. 

In   Fig.  96  the  additional  secant   plane  is  horizontal,  giving  a  vertical 


36 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


trace  VN,  and  cutting  the  plane  P  in  the  horizontal  {p'm',  o"in"),  and  the 
plane  M  in  the  horizontal  i^n'in',  n"m").  As  these  two  horizontals  lie  in 
the  same  plane,  they  intersect  in  the  point  {in' ,  m"),  and  hence  (a'jn', 
a"m")  is  the  line  of  intersection  required. 

In  Fig.  97  the  additional  secant  plane  is  vertical  in  position,  and  cuts 
the  given  planes  in  verticals  which  intersect  in  the  point  (;;/,  m"). 

In  the  first  case  the  line  of  intersection  lies  in  the  first  and  third 
dihedral  angles ;  in  the  second,  in  the  second  and  fourth  dihedral  angles. 

In  Fig.  98  the  auxiliary  secant  plane  is  any  plane  N  assumed 
within  the  limits  of  the  drawing.  The  intersections  of  this  plane  with 
the  given  planes  having  been  determined  as  in  Fig.  86,  the  point  in 
which  these  lines  intersect  is  a  point  in  the  line  of  intersection  sought. 

Fig.  99 


In  Fig.  99  the  auxiliary  secant  plane  is  a  new  vertical  plane  perpen- 
dicular to  GL,  giving  the  new  vertical  traces  b"m^  iand  c"m^.  Restoring 
the  point  ;/z/  to  the  primitive  planes  of  projection,  {a'ln' ,  a"m")  is  the 
line  of  intersection  sought. 

98.  In  Figs.  100  and  loi  the  two  intersecting  planes  are  parallel  to 
GL ;  hence,  when  cut  by  the  coordinate  planes,  give  lines  which  intersect 
at  infinity.      The  auxiliary  plane   may  be  any  plane   not  parallel  to   GL. 

Fig.  100  ^is- 10^ 


— ^ ~~Z^->'^\ 

^^^/>^             1 

/j^/^^^(^^7/ 

./-:.■/ 1 !     1 

V    i    ;    i/,^"' 

'M  ^i-7         HP 

\1/               H3. 

\ 


99.  In  Figs.  102  and  103  the  traces  do  not  intersect  within  the  limits 
of  the  drawing ;  hence  two  points  of  the  line  of  intersection  must  be 
determined  by  means  of  two  additional  auxiliary  planes. 

In    Fie.   102    a    horizontal   and  a  vertical   secant   plane   have  been  em- 


PARALLEL   AND  LNTERSECTING  PLANES. 


17 


ployed,  giving,  respectively,  the  pcjints  {o\  o")  and  (/>',  /")  through  which 
the  line  of  intersection  passes. 

In  Fig.    103  the  secant  planes  L  and  iV  are  parallel  to   GL,  having  a 
common    horizontal    trace    ML   and    cutting   each   upon   the   givep  planes 


a  line  the  vertical  projections  of  which  are  a'b',  c'b' ,  c'd',  f'd'.  The 
points  g'  and  //  in  which  they  intersect  determine  the  vertical  pro- 
jection g'h'  of  the  required  line  of  intersection. 

Two  additional  secant  planes,  having  a  vertical  trace  VQ  in  common, 
give  in  a  similar  manner  the  horizontal  projections  of  two  lines  whose 
points  of  intersection  determine  the  horizontal  projection  of  the  line  of 
intersection. 

100.  When  the  given  planes  are  three  in  number,  the  lines  of  inter- 
section must  either  be  parallel  or  intersect  in  a  common  point. 


Their  relative  positions  may  be  reduced  to  the  following  five: 
(i)  Parallel  to  one  another,  giving  no  intersection. 

(2)  Two  parallel,  mtersected  by  the  tiurd  ;   givmg  parallel  lines  of  in- 
tersection. 


38 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


(3)  Passing  through  a.  common  line  which  becomes  their  line  of  in- 
tersection. 

(4)  Intersecting  two  and  two  and  forming  a  prismatic  surface;  giving 
parallel  lines  of  intersection. 

(5)  Forming  a  trihedral  angle;  giving  lines  of  intersection  which  meet 
in  a  common  point. 

II.    THE   LINE  AND   PLANE. 

lOl.   Problem. —  To  find  the  piercing-point  of  a  line  on  a  given  plane. 

The  solution  consists  in  passing  through  the  given  line  any  auxiliary 
plane,  in  determining  the  line  in  which  this  plane  cuts  the  given  plane, 
and  in  finding  the  point  in  which  the  given  line  intersects  the  line  thus 
determined. 

Let  {VP,  HP)  be  the  given  plane  (Figs.  104,  105),  and  {a'b',a"b")  the 
given  line. 

The  construction  becomes  extremely  simple  when  the  auxiliary  secant 


plane  passing  through  the  line  is  assumed  to  be  the  projecting  plane  of 
the  line  itself, — in  Fig.  104  to  H,  and  in  Fig.  105  to  V  (Art.  'j']).  Find 
the  line  of  intersection  between  the  given  plane  and  the  projecting  plane, 


and  the   point  {o\  o"\  in  which  the  line  ab  in  space  intersects  it,  is  its 
piercing-point  on  the  given  plane. 

102.   In  Figs.   106  and    107  the  given  plane  is  perpendicular  to  either 
plane  oi  projection. 


THE   LINE  AND  PLANE. 


39 
and 


In   Figs.   io8  and    109  the  plane   is  given  by  three  points,  a,  b,  r, 
the  line  by  its  projections  {d'e',  d"e"). 

Connect  the  points  by  auxiliary  lines,  and  pass  through  the  given  line 
a  projecting  plane  (Art.  loi)  to  H,  as  in  Fig.  108,  or  to  /'  as  in  Fig. 
109,  cutting  the  connecting   lines  in  points  which    determine  the   line  of 


^-J>>' 


1^ 

-|--i — ! ^ 

1    >. 

a'k-i-- 

~r-t 1 

1          X 

^^^'"v-^Jo"!  .^^^^'''' 


Vie- 109 


grir^- 


intersection.  The  point  {0',  o")  in  which  this  line  intersects  the  given 
line  is  the  piercing-point  required. 

In  Figs,  no  and  iii  the  given  line  ab  lies  in  a  profile  plane.  In 
the  first  case  the  auxiliary  plane  has  been  passed  through  the  given  line 
(a'b',  a"b")  parallel  to  GL.  Its  intersection  with  the  given  plane  P  is 
found  by  the  use  of  additional  auxiliary  planes. 

Thus,  through  {a' ,  a")  conceive  a  horizontal  secant  plane  to  be  passed ; 
it  cuts  the    plane  P  in    the    horizontal  {c'e',  c"e"),  and   the  auxiliary  plane 


in  a  line  {c'cj',  c"a")  parallel  to  GL.  The  point  {c',  c")  in  which  these  two 
lines  intersect  is  a  point  in  the  line  of  intersection  sought. 

A  second  point  {g',  g")  may  be  determined  in  a  similar  way  by  means 
of  a  horizontal  secant  plane  through  {b' ,  b");  the  intersection  (<?',  o")  of 
ab  with  the  line  {c'g',  c"g")  connecting  these  two  points  is  the  piercing- 
point  required.  ^ 

In  Fig.  1 1 1  the  auxiliary  plane  is  the  profile  plane  ol  the  given  line. 
By  a  change  of  ground-line  its  intersection  /'V,'  with  the  plane  P,  and 
the  new  vertical  projection  a,'b^',  may  be  found,  and  with  them  o,' ,  the 
supplementary  projection  of  the  piercing-point  required. 


40 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


103.  Theorem, — A  right  line  perpendicular  to  a  plane  in  space  gives  pro- 
jections which  are  respectively  perpendicular  to  the  traces. 

In  Fig.  112  let  P  be  the  given  plane,  and  ab  a  line  perpendicular  to 
it.  The  plane  projecting  ab  upon  H  is  not  only  perpendicular  to  that 
plane,  but  also  to  the  given  plane  P;  hence,  being  perpendicular  to  two 
planes, "it  is,  by  Geometry,  perpendicular  to  their  line  of  intersection  HP, 
or  the  horizontal  trace.  From  this  it  follows  that  HP  is  perpendicular 
to  every  line  in  the  projecting  plane  which  passes  through  its  foot  on 
that   plane,  and  thus  to  a"b",  the  horizontal  projection  of   the  given  line. 


Fig.  113 


The  converse  of  this  proposition  is  likewise  true,  viz.:  if  the  projec- 
tions of  a  line  are  perpendicular  respectively  to  the  traces  of  a  plane,  the  li?ie 
is  perpendicular  to  the  plane. 

Fig.  113  represents  the  graphical  solution  of  the  case. 

104.  Problem. —  Through  a  ^iven  point  in  space  to  pass  a  plane  perpen- 
dicular to  a  given  line. 

Let  {a',  a")  be  the  given  point  (Fig.  114),  and  {b'c',  b"c")  the  given  line, 

By  the  conditions  of  the  problem  the  traces  of  the  required  plane 
must  be  respectively  perpendicular  to  the  projections  of  the  given  Hne. 
Through  a  lead  a  vertical  {a'd' ,  a"d!')  of  that  plane,  find  its  piercing- 
point  {d',  d"),  and  draw  the  traces  perpendicular  to  the  pi-ojections  of  the 
given  line. 


105.  Vko^UEU.— Through  a  given  line  to  pass  a  plane  perpendicular  to  a 
given  plane. 

Let  {a'b' ,  a"b")  be  the  given  hne,  and  ( VP,  HP)  the  given  plane  (Fig. 
115). 


THE  LINE   AND  PLANE. 


41 


From  any  point,  as  {c\  c"),  of  the  given  line  lead  a  perpendicular 
{c'd',  c"d")  to  the  plane,  find  the  piercing-points  {a' ,  a'-'),  {d';  d")  and 
(/,  /')  of  these  two  lines,  and  the  traces  of  the  required  plane  pass 
through  them. 

I06.  Problem. —  To  pass  a  plane  through  a  given  point  and  parallel  to 
tzvo  given  lines. 

Let  {a',  a")  be  the  given  point,  and  {b'e\  b"c"),  (d'e',  d"e")  the  given 
lines  (Fig.   ii6). 

Through  a  lead  lines  parallel  to  the  given  lines  and  find  their  pierc- 
ing-points;  the  traces  of  the  plane  pass  through  them. 


107.  Problem. —  To  project  a  given  point  upon  a  plane. 

Through  the  point  pass  a  line  perpendicular  to  the  plane  (Art.  103), 
and  find  its  piercing-point  upon  it. 

108.  Problexl — To  pass  through  a  given  point  a  line  zohieh  intersects  tzi'O 
given  right  lines. 

1st  Solution. — Through  each  line  and  the  given  point  pass  a  plane; 
their  line  of  intersection  is  the  line  sought. 

2d  Solution. — Determine  the  traces  of  a  plane  which  contains  one  of 
the  given  lines  and  the  point,  find  the  piercing-point  of  the  second  line 
upon  this  plane,  and  connect  the  point  so  found  with  the  given  point. 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


CHAPTER  V. 
I.    THE  GEOMETRICAL  SURFACES. 

109.  Every  surface  may  be  regarded  as  generated  by  the  movement 
of  a  line  either  fixed  or  variable  in  form,  and  regulated  in  its  change 
of  position  in  accordance  with  some  definite  law. 

Thus,  a  plane  may  be  described  by  the  motion  of  a  right  line  which, 
constantly  touching  a  second  right  line,  remains  parallel  to  its  original 
position ;  or,  again,  it  may  be  described  by  turning  the  first  line  around 
the  second  in  such  a  way  that  the  two  shall  remain  at  right  angles  to 
each  other  and  touch  in  a  fixed  point. 

The  moving  line  is  termed  the  generatrix,  and  the  line  which  guides 
the  movement  the  directrix. 

lOi.  Surfaces  may  be  divided  into  two  general  classes: 

(i)  The  rided  or  rectilinear,  or  such  as  may  be  generated  by  the  mo- 
tion of  a  right  line. 

(2)  The  curvilinear  proper,  or  such  as  admit  of  no  rectilinear  genera- 
trices. 

The  following  table  exhibits  these  classes  and  their  subdivisions: 

111.  The  plane  surfaces  are  such  as  are  limited  by  plane  faces,  as 
prisms,  pyramids,  etc. 

The  single-curved  surfaces  are  those  which  admit  of  rectilinear  genera- 
trices, and  are  of  three  kinds : 

(i)  Cylindrical  surfaces,  in  which  all  the  positions  of  the  generatrix 
are  parallel. 

(2)  Conical  surfaces,  in  which  all  the  positions  of  the  generatrix  ititer- 
sect  in  a  common  point. 

(3)  Surfaces  in  which  the  rectilinear  elements  intersect  two  and  two. 
The  double-curved  surfaces   are   those  on  which    no    right   line    can    be 

drawn,    as    spheres,    ellipsoids,    etc.,    and    which,    therefore,    can    only    be 
generated  by  curves. 

112.  Of  the  single-curved  surfaces  the  developable  are  those  in  which 
successive  rectilinear  elements  can  be  brought  in  contact  with  a  plane 
without  crumpling,  folding  or  tearing;  the  warped  are  those  in  which 
such  a  procedure  is,  in  general,  impossible. 


PLANE    SURFACES. 


43 


II.     PLANE    SURFACES. 

113.  The  prism  is  a  surface  generated  by  the  motion  of  a  right  line 
which,  while  remaining  parallel  to  its  original  position,  glides  along  the 
perimeter  of  any  polygon. 

It  niav  likewise  be  generated  by  a  polygon  which,  while  remaining 
panilkl  to  its  original  position,  has  one  of  its  points  in  constant  contact, 
during  its  motion,  with  a  given  line.  Thus,  the  directrix  of  the  first 
generation  may  become  the  generatrix  of  the  second,  and  reciprocally. 

The  surface  thus  determined,  if  limited  in  extension  by  planes,  has 
two  bases  and  an  altitude  equal  to  the  perpendicular  distance  between 
the  two. 

When  the  base  is  a  parallelogram,  the  prism  becomes  the  parallclo- 
pipcd ;    when  all  the  faces  are  squares,  the  cube. 

114.  A  Pyramid  is  a  surface  generated  by  the  right  line,  one  point  of 
which  remains  fixed  while  the  line  itself  glides  around  the  perimeter  of 
any  pol3'gon. 

Should  the  generatrix  extend  beyond  the  fixed  point,  termed  the 
vertex,  an  upper  and  a  lower  surface  will  be  formed,  the  limited  j)ortions 
of  which  are  determined  by  the  bases. 

The  lines  drawn  from  the  vertex  to  the  angular  points  of  the  base 
and   marking  the  intersections  of  the  adjacent  faces  are  termed  the  edges. 


e'     f   0'     h' 


^Y' kf 

d  11 Tc 


1      1 

1    1 

.7"  ir' 

ll 


a"  €' 


b"    d"  c" 


In  the  prism  the  edges  are  parallel;  in  the  pyramid  they  converge  to 
a  common  j)oint,  the  vertex. 

Both  surfaces  take  their  distinctive  names  from  their  bases,  as  tri- 
angular, hexagonal,  etc. 

115.  Proiu.KM. —  To  find  the  projections  of  a  parallelopiped  of  given  dimen- 
sions. 

Let  the  base  be  assumed  as  a  square  of  i  inch,  the  altitude  2  inches, 
and   let  the   prism    stand  on  //"  at  a  distance  of   i    inch   from  V. 

(i)  The  lower  base  resting  on  H  is  its  own  projection  (Fig.  117)  on 
that  plane;  hence,  draw  a"b"c"d"  a  square  of  the  given  dimensions,  and 
mark  its  vertical  projection  on  GL. 


44 


ELEMENTS  OF  DESCRIPTIVE    GEOMEIKY. 


(2)  The  axis  being  vertical,  so  are  the  edges ;  hence,  measure  above 
GL  a  distance  equal  to  the  altitude,  or  2  inches,  and  draw  a  horizontal 
line. 

(3)  The  bases  being  equal  and  parallel,  their  projections  will  respect-, 
ively  follow  the  same  conditions;  hence,  e'f'g'h'  is  the  vertical  projec- 
tion of  the  upper  base,  and  a'e' ,  d'f,  c'g',  b'h'  the  projections  of  the 
edges. 

116.  Fig.  118  represents  the  projections  of  a  cube  whose  faces  incline 
to   V  and  are  perpendicular  to  H. 

Fig.  1 19  represents  the  projections  of  a  right  pentagonal  prism  whose 
bases  are  parallel  to  V,  and  whose  edges,  as  a  consequence,  are  parallel 
to  H. 

117.  Problem. —  To  project  a  right  pyramid  of  given  dimensions. 

Let  the  base  be  a  hexagon  of  \\  inches,  the  altitude  3  inches,  and 
the  pyramid  stand  on  //,   i   inch  distant  from    V. 

(i)  Measure  i  inch  below  GL,  and  draw  the  side  a"b"  of  the  hexagon 
equal  to  \\  inches.  Construct  the  hexagon  on  this  line  as  a  base,  and 
find  its  vertical  projection  in  GL  (Fig.  120). 

(2)  The  axis  being  vertical  and  passing  through  the  centre  of  the 
base,  v"  is  its  horizontal  projection,  and  v'v'-,-  equal  to  3  inches,  its  ver- 
tical projection. 


(3)  The  upper  extremity  of  the  axis  is  the  vertex ;  hence  the  edges 
are  determined  by  joining  its  projections  {v\  v")  with  those  of  the  angu- 
lar points  of  the  base. 

Fig.  121  represents  the  projections  of  the  double  octagonal  pyramid 
from  which  the  solid  angles  have  been  cut  by  planes  parallel  to  the 
common   base. 

118.  Among  the  plane  surfaces  are  the  regular  polyhedrons,  or  those 
in  which  the  sohd  angles  are  equal.  They  are  five  in  number,  viz.:  the 
tetrahedron,  in  which  the  solid  angles  are  bounded  by  three  equilateral 
triangles;  the  cube,  in  which  they  are  bounded  by  three  squares;  the 
octahedron,  by  four  equilateral  triangles  ;  the  dodecahedron,  by  three  pen- 
tagons;    and  the  icosaJicdron,  bv  five  equilateral  triangles. 


PLANE    SURFACES. 


45 


119.   Problem. —  To  determine  the  projections  of  the  octahedron. 

Assume  that  one  face  lies  in  H  with  one  side  parallel  to  GL  (Fig.  122). 

On  the  given  side  {a'b' ,  a"b")  construct  an  equilateral  triangle.  As  a 
square,  abde,  whose  sides  form  the  common  base  of  the  two  pyramids  lies 
in  a  plane  parallel  to  GL,  its  projection  on  a  new  vertical  plane  G,L,  is 
a  line  equal  to  the  given  side,  and  forming  at  its  middle  point  a  right 
angle  with  the  projection  of  the  axis  on  that  plane.  Hence  from  a^  as  a 
centre  describe  an  arc  with  a  radius  equal  to  one  half  of  the  given  side; 
the  line  <:,'//  is  the  new  vertical  projection  of  the  axis,  and  a^b^d^e^  that 
of  the  base,  from  which  a"b"d"e"  may  be  readily  determined. 

As  the  faces  diagonally  opposite  are  parallel  to  each  other,  the  upper- 
most one  is  parallel  to  H \  hence  {d"e"f")  and  {d'e'f)  are  its  horizcjntal 
and  vertical  projections  respectively.  Join  the  vertices  (t',  c")  {f\f")  with 
the  angular  points  of  the  base  and  complete  the  surface. 


120.   Problem. —  To  deterviine  the  projections  of  the  icosahedron. 

Assume  the  axis  of  the  surface  to  be  vertical  (Fig.   123). 

On  a  given  side  {a"b",  a'b')  construct  a  regular  pentagon  {a"b"c"d"e'\ 
a'b'i'd'e'),  having  one  radius  parallel  to  GL. 

As  the  upper  portion  of  the  surface  is  a  regular  pyramid,  joining  m" 
with  the  angular  points  of  the  pentagon  determines  its  horizontal  projcc- 
tion.  To  determine  the  vertical  projection  make  d'm'  equal  to  the  given 
side,  since  this  edge  of  the  pyramid  is  parallel  to   V. 

Construct  a  second  pentagon,  equal  to  and  concentric  with  the  first. 
3ut  in  such  a  way  that  its  angular  points  f"g"h"i"k"  shall  lie  in  the  middle 
-y{  the  arcs  which  are  subtended  by  the  sides  of  the  first  pentagon.  Join 
he  vertex  {n'\  n')  with  each  of  the.se  points  to  form  the  lower  pvramid 
)f  the  surface,  and  from  each  angular  pomt  of  the  upper  pentagon  draw 


46 


ELEMENTS   OF  DESCRIPTIVE   GEOMETRY. 


lines  to  the  two  adjacent  points  of  the  lower  pentagon,  thus  determining 
the  intermediate  zone  between  the  two  pyramids. 

As  c"h"  is  by  this  construction  parallel  to  F,  the  vertical  projection 
c'h'  is  shown  in  its  true  length  ;  hence,  the  plane  of  the  lower  pentagon 
is  fixed  by  making  the  distance  from  c'  to  h'  equal  to  the  given  side. 

121.  As  polyhedra  are  represented  by  the  projections  of  the  edges 
which  limit  the  faces,  and  as,  by  the  preceding  examples,  it  is  manifest 
that  all  the  faces  are  not  visible,  it  will  be  well,  just  here,  to  consider  in 


what  way  the  projections  are  affected  and  the  means  by  which  they  ma^^ 
be  represented. 

The  point  at  which  the  eye  is  located  in  space  is  the  point  of  sight 
If  its  distance  from  the  object  be  finite,  the  lines— termed  visual  rays- 
drawn  from  the  eye  to  each  point  of  that  object  will  be  divergent,  and 
the  projections,  though  similar  to  the  original  in  respect  of  form  and 
position,  will  be  larger  or  smaller  as  the  object  is  placed  before  or  behinc 
the  plane  of  projection  (Fig.   124). 

In  orthographic  projection,  if  the  projecting  lines  be  regarded  as  visua 


PLANE   SURFACES. 


A1 


rays  (Fig.  125),  the  point  of  sight  must  be  at  an  infinite  distance  in  order 
that  their  parallelism  may  be  established.  Whence  it  follows  that  the 
horizontal  projection  is  that  view  of  an  object  which  is  obtained  at  an 
infinite  distance  from  the  horizontal  plane,  and  the  vertical  projection  is 
that  view  of  an  object  which  is  obtained  at  an  infinite  distance  from  the 
vertical  plane. 

122.  In  the  vertical  projection,  therefore,  those  parts  of  the  same  ob- 
ject or  of  any  other  object  which  lie  to  the  front  may  cover  or  conceal 
those  which  lie  to  the  rear ;  and,  in  a  similar  way,  the  upper  parts  of 
the  same  or  any  other  object  may,  in  the  horizontal  projection,  cover  or 
conceal  the  lower  parts.  The  solution  of  these  difficulties  will  be  mate- 
rially simplified  by  the  observance  of  the  following  facts : 

(i)  The  outlines  or  extreme  limiting  lines  of  both  projections  are 
always  visible. 

(2)  Only  that  part  of  the  vertical  projection  can  possibly  be  concealed 
which  is  covered  by  some  other  part  in  the  same  projection.  The  hori- 
zontal projection  of  the  parts  vertically  in  covimon  will  determine  the 
question. 

(3)  Only  that  part  of  the  horizontal  projection  can  possibly  be  con- 
cealed which  is  covered  by  some  other  part  in  the  same  projection.     The 


Kig.  12^1 


Fie.  li2C 


vertical  projection  of  the  parts  Jiorizontally  in  common  will  determine  the 
question. 

(4)  In  passing  from  a  part  seen  to  .a  part  concealed,  the  outline  of  the 
projection  must  be  crossed. 

123.    Figs.   126  and   127  will  serve  as  illustrations. 

(i)  In  Fig.  126  the  outline  of  the  combined  projections  of  the  three 
solids  has  been  drawn  in  full  lines,  in  accordance  with  the  first  principle. 

(2)  An  inspection  of  the  horizontal  projection  shows  that  a  part  of 
each  surface  is  included  within  the  limits  of  this  outline.  It  is  manifest, 
however,  from  an  examination  of  these  same  parts  in  the  vertical  projec- 
tion, that  the  upper  base  of  the  triangular  prism  rises  above  the  other  two 
surfaces ;  hence,  in  looking  down  upon  //,  the  prism  will  conceal  such  parts 
of  them  as  lie  within  its  contour.  In  like  manner  it  conceals  one  side  of 
the  lower  base  {a'b\  a"b")  and  the  under  face  ia'b'c'd',  a"b"c"d")  (Fig.  127). 

(3)  An    inspection  of   the  vertical   projection   (Fig.   126)   shows   that   a 


48 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


part  of  each  surface  is  included  within  the  general  contour;  but  by  an 
examination  of  the  horizontal  projection  it  is  evident  that  the  cylinder 
lies  to  the  front  of  the  other  two  surfaces,  and  hence  covers  and  con- 
ceals all  those  parts  of  the  other  surfaces  which  lie  within  its  limits. 
It  will  also  be  seen  that  the  pentagonal  prism  covers  a  part  of  the  tri- 
angular prism. 

The  dotted  lines  indicate  the  hidden  parts. 


Fig.ise 


iTis.iar 


L v^-^ 

// 

A  \ 

/        1                       s 

\  / 

/      '             / 

V 

/     1 

/ 

i 

124.  When  the  methods  of  Descriptive  Geometry  are  applied  to  ma- 
chiner)^  architecture,  etc.,  the  term  p/a?t  is  employed  to  denote  the  hoi'i- 
zontal  projection,  and  elevation  the  vertical  projection ;  front  elevation 
referring  to  an  ordinary  vertical  projection,  and  side  elevation  to  that 
which  is  determined  by  means  of  a  profile  plane. 

125.  Problem. —  To  project  an  oblique  pyramid  of  given  dimensions. 

Let  the  pyramid  (Fig.  128)  be  pentagonal,  the  radius  of  circumscrib- 
ing circle  f  of  an  inch,  the  altitude  2  inches,  the  axis  inclining  to  the 
base  at  an  angle  of  45°  to  the  left;  and  let  it  stand  on  H,  i  inch  from  V. 

(i)  Place  the  surface  in  its  simplest  position  with  its  axis  parallel  to 
V.     Find  the  projections  of  the  base  (Art.  117). 

(2)  The  axis  being  parallel  to  V,  find  its  projection  on  that  plane,  and 
determine  its  horizontal  projection  parallel  to  GL. 

(3)  Join  the  upper  extremity  or  vertex  {v\  v")  with  the  respective 
angular  points  of  the  base,  observing  that  for  the  vertical  projection  the 
edge  {a'v',  a"v")  at  the  back  of  the  surface  is  concealed,  and  for  the  hori- 
zontal projection  the  edge  {b'v\  b"v")  and  two  sides  of  the  base  {a"d",  a'b') 
{b"c",  b'c')  on  the  under  part  of  the  surface  are  concealed. 

126.  Problem. —  To  project  an  oblique  prism  of  given  dimensions. 

Let  the  dimensions  be  as  in  the  preceding  case  (Fig.  129),  but  the  axis 
parallel  to  H  and  inclining  to  V  towards  the  right  hand. 


SINGLE-CUR  VED   SUKEA  CES. 


49 


(i)  Place  the  prism  with  its  bases  parallel  to  V,  and  determine  the 
vertical  projection  of  the  one  nearer  to  that  plane  ;  a"d"  is  its  horizontal 
projection. 

(2)  The  axis  beinj^  parallel  to  //,  determine  its  horizontal  projection 
o"o"  of  the  given  length  and  in  the  required  position  ,  o'o'  is  its  vertical 
projection. 

(3)  The  two  bases  being  parallel,  the  projections  are  likewise  parallel: 
hence   the   vertical    projection   of    the    second    base    has   its   homologous 


inigvaao 


a'y 

^P'-f'  , 

^\ 

\ 

Y  !    c'l  1       ' 

\--'} 

I         1 

sides   parallel   to   the    first,    and   its   horizontal    projection   a    line    parallel 
to  GL. 

(4)  Join  the  corresponding  angular  points  of  the  bases  to  determine 
the  edges. 

III.    SINGLE-CURVED   SURFACES. 

127,  The  cylinder  is  a  surface  generated  by  the  motion  of  a  right 
line  which  constantly  touches  the  perimeter  of  any  curve  and  remains 
parallel  to  its  original  position.  The  different  positions  of  the  generatrix 
indicate  the  rectilinear  elements  of  the  surface. 

The  surface  may  also  be  generated  by  the  motion  of  a  curve  in  such 
a  way  that  each  point  shall  describe  a  line  parallel  to  a  given  right 
line.  The  different  positions  of  the  curve  indicate  the  curvilinear  elements. 
The  same  distinctions  as  to  bases,  axes,  etc.,  are  to  be  noticed  as  in  the 
case  of  the  prism,  an  analogy  at  once  apparent,  as  the  curve,  whether 
regarded  as  generatrix  or  directrix,  may  be  considered  as  a  polygon  <^f 
an    infinite   number  of  sides. 

Cylinders  are  named  according  to  the  character  of  the  base,  as  cir- 
cular, elliptical,  parabolic,  etc. 

128.  The  cone  is  a  surface  generated  by  the  motion  of  a  right  line 
which  passes  through  a  fixed  point,  the  vertex,  and  constantlv  touches 
the   perimeter  of  any  curve,      if    the    generatrix  extend   bevond   the  fixed 


50 


ELEMENTS   OF  DESCRIPTIVE   GEOMETRY. 


point  it  will  describe  two  branches  of  the  surface,  one  on  either  side  of 
the  vertex,  termed  the  upper  and  lower  nappes.  The  different  positions  of 
the  generatrix  indicate  the  rectilmear  elements  of  the  surface. 

The  cone  may  also  be  generated  by  the  motion  of  a  curve  in  such  a 
way  that  any  point  in  its  plane  shall  move  along  a  right  line  passing 
through  a  fixed  point,  and  the  curve  vary  in  proportion  to  its  distance 
from  that  point.  The  different  positions  of  the  curve  indicate  the  eurvi- 
linear  elements  of  the  surface. 

If  the  vertex  of  the  cone  be  assumed  at  a  maximum  distance,  the 
surface  becomes  cylindrical ;  if  at  a  minimum  distance,  a  plane.  If  the 
radius  be  infinitely  increased,  the  surface  becomes  a  plane ;  if  infinitely 
diminished,  a  right  line. 

129.    Problem. —  To  project  a  right  cone  of  given  dimensions. 


The  solution  is  precisely  similar  to  that  of  the  right  pyramid,  the- 
base  of  the  cone  being  regarded  as  a  polygon  of  an  infinite  number  of 
sides;  but  as  the  edges  now  become  the  rectilinear  elements  of  the  sur- 
face, only  those  appear  in  the  projections  (Fig.  130)  which  lie  in  the 
extreme  projecting  planes  to  either  plane  of  projection.  The  same  holds- 
true  of  the  right  cylirder  (Fig.  131). 

130.    Problem. —  To  project  an  oblique  cone  of  given  dimensions. 


Fig.  13S 


The   solution  is  similar  to  that  of  the  oblique  pyramid,  with  this  dif- 


WARPED   SURFACES. 


51 


ference :  that,  there  being  no  distinctive  elements  or  edges  on  the  surface, 
th(jsc  only  are  represented  which  form  the  extremes  or  tangents  to  the 
]>rojections  (Fig.   132). 

The  same  holds  true  of  the  oblique   cylinder  (Fig.   133). 

131.  The  loarpcd  surfaces  are  those  which  are  generated  by  the 
right  line,  but  cannot  be  brought  in  ctjntact  or  spread  up(jn  a  jjlane 
without  hjlding  or  tearing. 

The  simplest  of  these  surfaces  is  the  Jiypcrboloid  of  one  nappe,  which 
is  generated  by  the  motion  of  a  right  line  which  constantly  touches 
three  directrices  (Fig.   134). 

Thus,  let  {a'b\  a"b")  be  a  given  right  line  and  {f'd\  e"d")  a  vertical 
axis  around  which  the  given  line  is  tQ  revolve  in  such  a  way  that  its 
two  extremities  and  the  intermediate  point  (/,  e")  shall  glide  upon  the 
perimeters  of  three  given  circles.  By  these  conditions  the  rectilinear 
generatrix   preserves  the  same  relative  position  to  and  distance  from  the 


axis;  hence  its  horizontal  projections  a"b",  f"g'\  etc.,  must  be  tangent 
to  the  smallest  circle,  the  circle  of  the  gorge,  on  which  the  point  e  glides. 
As  the  upper  and  lower  extremities  lie  in  the  larger  circles,  the  ver- 
tical projections  are  readily  determined  by  means  oi  ordinates. 

132.  The  Inperbolie-paraboloid  or  xcarped  plane  is  a  surface  (Fig.  135) 
which  is  generated  by  the  motion  of  a  right  line  which  glides  upon  two 
right  lines  not  in  the  same  plane,  and  remains  constantly  parallel  to  a 
given  plane,  termed  the  plane  directer. 

Thus,  let  {a'b',  a"b")  and  {e'd',  e"d")  be  the  two  right  directrices,  and 
let  //  be  the  plane  directer.  Any  plane  parallel  to  H  cuts  the  two 
lines  in  points  which  determine   the  position  of  the  generatrix. 

133.  The    rig/it   eonoid  is   a    surface    generated    bv   a    risjht    line   which 


52 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


glides  •  along  a  curve    in    such    a  way  that   it   constantly   touches   a   right 
line  and  remains  parallel  to  a  plane  directer. 

Thus,  let  {a'e'c' ,  a" c" c")  be  the  given  curve  (Fig.   136),  and  {b' d' ,  b" d") 
be  the  given  line,  and  let  H  be  the  plane  directer. 


Any  horizontal  plane  cuts  the  curve  and  line  in  points  which  deter- 
mine the  generatrix  in  position. 

134.  The  Jielicoids  are  surfaces  generated  by  the  right  line  which 
glides  along  a  helix,  and  maintains  an  invariable  position  to  the  axis  of 
Ihe  curve. 

Tne   helix   is   a   curve   generated    by  a   point  which    moves   along   the 


Fig.    138 


Kig.  137- 

d 

6' 

-j.---4^--i.- 

-] -I S-T 

.-l^. 1 L- 

:i:i;pfj^ 

^^f^":~i: 

->---!—{- 

tSi. 

d 

W:- 

Fig 

139 

Sv|       1 \^\ 

!--w^-^^-i 

ppi 

%-\ 

!  p^ 

Mwi 

U-h-/-- 

"v^ 

Jr" 

^-Ll 

surface  of  a  cylinder  in  such  a  way  that  a  constant   ratio   is  maintained 
between  the  measure  of  its  rotation  and  ascent. 

Thus,  let  {a'a',  a")  be  the  element  of  the  cylinder,  and  let  its  foot  be  the 
moving  point  (Fig.  137).  Divide  the  circumference  of  the  base  into  eight  and 
the  height  into  sixteen  equal  parts,  and  let  it  be  assumed  that  for  every 
turn  of  an  eighth  around  the  axis  the  point  {a',  a")  ascends  one  sixteenth 


'  DOUBLE-CURVED   SURFACES  AND   SURFACES  OF  REVOLUTIOX.  5 J 

of  the  heii^ht.  Then  when  the  element  has  moved  into  the  position  {b'b\ 
b")  the  point  will  have  risen  to  {b' ,  b"),  and  when  the  third  position 
{c'i-\  c")  has  been  attained  to  {c',  c"),  and  so  on  in  regular  succession 
until  two  complete  turns  have  been  made. 

135.  Pkoijlem. —  To  project  a  icarpcd  hclicoid. 

There  are  two  surfaces  belonging  to  this  class  which  are  to  be  noted: 
one  in  which  the  generating  line  intersects  the  axis  at  right  angles,  and 
another  in  which  it  intersects  it  at  an  invariable  acute  angle. 

Let  the  second  case  be  assumed  (Fig.  139),  and  let  {a'c\  a"c")  be  the 
initial  position  of  the  generatrix.  As  the  generatrices  are  all  equally 
inclined  to  the  axis,  the  distance  between  their  extremities  for  any  posi- 
tion will  be  constant ;  hence,  to  determine  a  sec(3nd  generatrix  {d'c' , 
d"e")  for  a  second  point  {d',  d")  of  the  helix,  lay  off  on  the  axis  the 
distance  n'c'  equal  to  o'c'.  Proceed  in  a  similar  manner  for  the  remain- 
ing positions  of  the  generatrices. 

IV.    DOUBLE-CURVED   SURFACES   AND   SURFACES   OF    REVOLUTION. 

136.  A  surface  of  revolution  is  generated  by  any  line  which  revolves 
arcjund  a  right  line,  termed   the  axis  of  rotation. 

Wiien  tiie  generatrix  is  plane  it  is  termed  a  meridian  line,  and  its- 
plane  a  meridian  plane.  During  rotation  every  point  of  the  meridian 
describes  the  circumference  of  a  circle  whose  plane  is  perpendicular  to 
the  axis  and  whose  centre  lies  in  it.  These  circles  arc  the  paralhls,  and 
may  also  be  generatrices  of  the  surface. 

137-  While  all  the  meridians  are  equal,  the  parallels  vary  in  size  ac- 
cording to  the  nature  of  the  generating  line.  When  the  meridian  becomes 
a  right  line,  tlie  surface  is  either  the  right  cylinder  or  cone  as  the  genera- 
trix is  parallel   to  or  intersects  the  axis. 

138.  Among  the  geometrical  surfaces  of  revolution  the  following  are 
i(}  be  noted : 

(i)  Ellipsoid  of  revolution,  which  is  generated  bv  the  complete  revo- 
lution of  a  semi-ellipse  arounvi  its  major  or  minor  axis,  giving.  resi)ect- 
ively,  the  prolate  and  oblate  ellipsoids. 

(2)  Hyperboloid  of  revolution,  which  is  generated  bv  the  com|>lete 
revolution  of  a  semi-hyj^erbola  around  one  of  its  axes.  It  is  indefinite 
in  extent,  being  limited  for  constructive  purjioscs  bv  jilanes  jierpendicular 
to  the  axis. 

(3)  Paraboloid  of  revolution,  which  is  generated  by  the  complete 
revolution  of  a  semi-parabola  around  an  axis  of  the  curve.  It  is  also 
indefinite  in  extent  and  limited  as  in  the  j^receding  case. 

(4)  Sphere,  generated  by  the  semicircle. 

139.  The  projections  of  surfaces  of  revolution  are,  in  general,  deter- 
mined by  the  meridians  and  parallels. 

The  Case  is  much  simplified  sh»)uUl  the  axis  of  the  surface  be  taken 
perpendicular  to  //.  In  such  a  position  the  vertical" projection  is  a  full 
meridian,    and    the    horizontal    projection    a    series   of    concentric    circles 


54 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


whose  number  will  depend  upon  the  important  parallels  characteristic  of 
the  surface. 

140.  In  addition  to  the  surfaces  of  revolution  the  following  are  to 
be  noted  : 

(i)  Tri-axial  ellipsoid  (Fig.  140),  which  is  generated  by  an  ellipse  aba'b' 
which  moves  in  such  a  way  that,  while  it  remains  parallel  and  similar 
to  itself,  its  centre  c  glides  along  the  axis  AA' ,  and  the  extremities  of 
the  major  axis  describe  the  ellipse  aAa'A'  at  the  same  time  that  the  ex- 
tremities  of  the  minor  axis  b  and  b'  describe  the  eUipse  bAb'A'. 

When  BD  equals  CD  the  ellipsoid  becomes  one  of  revolution,  and 
when  £D  —  CD  =  DA  it  becomes  the  sphere. 


Fig.  142 


(2)  Hyperboloid  of  one  nappe  (Fig.  141)  with  three  unequal  axes,  Avhich 
is  generated  by  an  ellipse  BaB'a'  which  moves  in  such  a  way  that,  while 
it  remains  parallel  and  similar  to  itself,  its  centre  glides  along  the  axis 
AA\  and  the  extremities  of  either  axis  have  a  given  hyperbola  BCB'C 
for  a  directrix. 

When  the  axes  are  equal  the  surface  is  one  of  revolution. 

(3)  Elliptic  paraboloid  (Fig.  142),  which  is  generated  by  an  ellipse  BaB'a' 
which,  while  remaining  parallel  and  similar  to  itself,  moves  so  that  the 
extremities  of  either  axis  have  the  parabola  BAB'  for  a  directrix. 

When  the  axes  are  equal  the  surface  is  one  of  revolution. 


SUPPLEMENTARY  PLANES  AND  PROJECTIONS.  55 


CHAPTER  VI. 
SUPPLEMENTARY   PLANES   AND   PROJECTIONS. 

141.  It  frequently  happens  in  practice — and  illustrations  are  not  want- 
ing in  the  preceding-  articles — that,  in  addition  to  the  coordinate  planes, 
other  ways  and  means  must  be  devised  to  render  the  graphical  S(jlution 
of  a  problem  possible.  The  change  in  the  position  of  the  ground-line  is 
an  expedient  of  this  character.  The  necessity  for  such  special  construc- 
tions arises  from  some  special  arrangement  of  the  objects  in  space  with 
reference  to  the  coordinate  planes. 

142.  Under  such  conditions  the  operations  may  be  simplified  in  two 
ways:  either  by  means  of  additional  planes  of  projection  or  by  some  neio  ar- 
rangement of  the  objeets  themselves.  In  other  words,  the  questions  to  be 
resolved  are: 

(0  Having  the  primitive  projections  of  any  object,  to  determine  new 
or  supplementary  projections  on  additional  planes. 

(2)  Having  the  primitive  projections  of  any  object,  to  determine  its 
projections  after  a  change  in  its  oivn  position  to  the  primitive  coordinate 
planes. 

The  discussion  of  the  first  condition  is  the  subject-matter  of  this 
chapter;    the  second  will  be  treated  in  a  succeeding  chapter. 

143.  The  employment  of  an  additional  plane  in  nowise  affects  the 
position  of  the  object  to  the  primitive  planes;  its  sole  end  is  to  present 
some  new  viczv  of  that  object,  thereb}'  admitting  of  a  clearer  comprehen- 
sion of  its  character  and  lessening  the  constructive  difficulties  of  the  case. 

144.  The  position  of  the  supplementary  plane  must  mainly  depend 
upon  that  of  the  object,  although,  wherever  practicable,  it  is  desirable 
that  it  should  be  as  simple  as  possible;  hence,  perpendicular  to  both  co- 
ordinate planes,  or,  at  least,  perpendicular  to  one  of  those  planes. 

145.  Whatever  position  may  be  adopted  for  the  supplementary  plane, 
the  projections  are  determined  precisely  as  in  the  case  of  the  primitive 
planes;  that  is,  by  letting  fall  from  the  object  perpendiculars  to  that  plane, 
and  by  marking  their  intersections  thereon. 

146.  The  constructions  thus  made  are  brought  into  the  drawing  plane 
by  revolving  the  supplementary  plane  aroimd  either  trace  into  either  of 
the  coordinate  planes. 

Thus  with  5  the  supplementary  plane  perpendicular  to  both  coor- 
dinate planes  (Fig.    143),  it  may  be  revolved  : 

(i)  Around  VS,  termed  the  vertical  supplementary  ground-line,  into  V 
(Fig.   144). 


56 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


(2)    Around   HS,  the  horizontal  supplementary    oround-line,  into   H  (Fig 

I45> 

147.    The  results  of  these  constructions  demonstrate  that: 

(i)   The    vertical    trace   or    ground-line    VS  (Fig.    144)    remams    fixed, 

while  the  horizontal  ground-line  HS  rotates  through   a  quadrant  until  it 

coincides  with  GL. 

(2)  All  distances  from  H  are  measured  on  5  at  right  angles  to  HS. 

(3)  All  distances  from   V  are  measured  on  5  at  right  angles  to   VS. 


Fig.  143 


Fig.  144 


VS 


"k A- 


r 


if. 

I — •- 

rs 

\ 

\ 
\ 

1 
1 
1 

^ 

HS 

\HS 


(4)  The  projecting  lines  to  5  are  represented  on  the  coordinate  planes 
by  ordinates  perpendicular  respectively  to  the  supplementary  ground-lines, 

148.  With  S  perpendicular  to  either  plane  of  projection  and  inclining 
to  the  other,  the  supplementary  ground-Unes  assume  the  positions  indi- 
cated in  Art.  68.  In  such  a  case  5  is  ordinarily  revolved  around  the 
inclining  ground-line  into  the  primitive  plane  in  which  that  hne  lies. 

Thus,  in  Fig.  146  5  has  been  revolved  around  VS  into  V,  and  in  Fig. 
147  around  HS  into  H. 


mg.T.AT' 


149.    The  results  of  this  rotation  demonstrate  that — 

(i)   The  inclining  ground-line  remains  fixed  during  the  rotation. 

(2)  As  the  relative  position  of  the  supplementary  ground-lines  is  not 
altered  by  the  rotation,  they  still  retain  a  rectangular  position  to  each 
other. 

(3)  The  projecting  lines  let  fall  from  the  object  upon  5  are  repre- 
sented by  the  ordinates  drawn  from  the  primitive  projections  perpen- 
dicular to  the  supplementary  grcjund-lines. 


SUPPLEMENTARY  PLANES  AND  PROJECTIONS. 


57 


(4)  The  intersections  of  these  ordinates  determine  the  required 
supplementary  projections. 

150.  When  5  is  made  to  incline  to  both  coordinate  planes — a  position 
least  calculated  to  insure  simplicity  of  construction — the  projections  of  the 
object  will  still  be  found  by  letting  fall  perpendicular  projecting  lines 
upon  that  plane.  As  in  the  preceding  cases,  the  ordinates  must  neces- 
sarily be  perpendicular  to  the  respective  ground-lines. 

151.  The  supplementary  plane  is  rotated  away  from  the  object  in 
order  to  avoid  the  confusion  which  would  necessarily  arise  from  the 
superiniposition  of  the  projections. 

152.  Problem. —  The  traces  of  t%i.'o  planes  being  givoi,  to  find  their  line 
of  intersection  by  tneans  of  a  profile  supplementary  plane. 

Let  {VO,  HO)  and  {VP,  HP)  be  the  given  traces  (Fig.  148),  and  let  5 
be  revolved  into    / '. 

The  points  in  which  HO  and  HP  cut  the  ground-line  HS  are  carried 
with  that  line  into  GL,  while  the  points  in  which  VO  and  /'/*  cut  VS 
remain  fixed;  hence  SO  and  SP  are  the  supplementary  traces,  and  their 
]>oint  of  intersection  a^'  the  supplementary  projection  of  the  line  of  inter- 
section. As  the  distances  from  the  supplementary  ground-lines  measure 
the  distances  from  the  coordinate  planes,  aa^'  is  the  height  above  //,  and 
fia^'  the  distance  behind  V.  The  intersection  is  effected  in  the  second 
angle,  and  {a'b' ,  a"b")  are  the  projections  sought. 

Kig.  140 

VS 


153.  Prorlkm. —  To  determine  the  oblique  supplementary  projection  of  a 
circle,  its  primitive  projections  being  given. 

Let  {VS,  HS)  be  the  supplementary  plane,  and  {a' . .  b' ..d\  a" . .  b" . .  d") 
the  given  circle. 

Fn^m  any  number  of  points  of  the  circle  (Fig.  149)  let  fall  projecting 
lines  upon  5,  which  lines  being  perpendicular  to  that  plane  are  projected 
in  ordinates  perpendicular  respectively  to   VS  and  HS. 


58 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


The  points  in  which  the  ordinates  cut  HS  are  revolved  with  that 
line  into  the  new  position  HS„  and  determine  the  several  ground-points. 
The  ordinates  drawn  from  these  points  intersect  those  drawn  from  the 
vertical  projection  in  \a"' ,  b'" ,  c'" ,  d"'\  through  which  the  required  pro- 
jection of  the  circle  passes. 

154.  Problem. —  To  find  the  primitive  projections  of  a  right  prism,  its 
supplementary  projection  being  given. 

Let  (KS,  HS)  be  the  given  plane  (Fig.  150),  and  {a!"  .  .  .  Ji'")  the  sup- 
plementary projection  of  the  prism. 

Assuming  the  prism  to  be  right,  the  axis  must  be  parallel  to  GL,  and 
hence   both   the   edges   and   their   projections.     As  the  distances   from   V 


are  measured  from  VS  in  the  supplementary  plane,  or  from  F5,  in 
the  revolved  position,  the  horizontal  projection  may  be  determined 
immediateh'  by  ordinates  drawn  from  the  supplementary  projection 
parallel  to  GL. 

And,  again,  as  the  distances  from  HS  measure  the  heights  above  H, 
parallels  to  HS  drawn  through  the  points  of  the  supplementary  projec- 
tion mark  these  distances  upon  KS, ;  whence,  upon  restoring  5  to  its 
original  position,  they  are  transferred  to  VS  and  determine  the  vertical 
projection. 


CHANGE   OF  POSITION  BY  ROTATION  AND  RABATTEMENT. 


59 


CHAPTER  VII. 

CHANGE    OF    POSITION    BY    ROTATION    AND    RABATTEMENT. 

155.  Instead  of  employing  the  method  of  the  preceding  chapter  the 
same  end  may  be  attained  by  changing  the  position  of  the  object  itself, 
thus  affording  new  views  and  consequently  new  projections. 

The  operations  by  which  this  is  accomplished  are  either  movements 
parallel  to  a  rectilinear  or  plane  directer,   or  movements  of  rotation. 

156.  In  these  changes  it  is  to  be  remembered  that  the  principal  ob- 
ject  is  to  facilitate  the  work  of  solution  ;  hence,  whatever  the  alterati(jn 
in  position,  the  simplification  of  the  constructions  must  be  kept  constantly 
in  view. 

157.  With  the  movement  of  rotation  there  are  necessarily  implied : 
(i)   An   axis   around    which   the    object    revolves — the   axis  of  rotation 

^i5(Fig.  151). 

{2)  The  fixed  distances  of  each  point  of  the  object  from  the  axis 
during  the  entire  rotation — the  radii  of  rotation  CD. 

(3)  The  foot  of  the  radius,  always  marking  a  point  in  the  axis — the 
centre  of  rotation  C. 

Kig.lCl 


(4)  The  locus  of  each  point — the  circle  of  rotation  D^DJ)^. 

{5)  The  plane  of  that  circle,  always  at  right  angles  to  the  axis — the 
plane  of  rotation. 

(6)  The  arc  D,D^  through  which  any  point  Z>,  revolves,  giving  the 
measure  of  rotation  D,CD^  for  every  other  point  of  the  object. 

Taken  together,  these  constitute  a  system  of  rotation  the  position  of 
■which  must   evidently  depend  upon  that  of  the  axis. 


6o 


LLEMEiYTS   OF  DESCRIPTIVE    GEOMETRY. 


158.  There  are  two  general  positions  which  such  an  axis  ma)^  assume 
to  the  moving  object:  it  may  have  one  or  more  points  in  common  with 
it,  or  may  lie  wholly  outside  it. 

In  connection  with  the  first  case  the  more  ordinary  positions  of  the 
axis  are: 

(i)    With  a  line,  it  has  a  point  in  common. 

(2)  With  a  plane,  a  point  or  line  in  common. 

(3)  With  a  plane  figure,  it  coincides  with  an  axis,  diameter  or  side, 
or  with  any  tangent  to  it. 

(4)  With  a  surface,  it  coincides  with  an  axis,  element  or  tangent. 

159.  In  the  second  case  the  ordinary  positions  of  the  axis  are : 

(i)    With  a  line,  it  is  either  parallel  to  it  or  lies  in  another  plane. 

(2)  With  a  plane,  it  is  parallel  to  it. 

(3)  With  a  plane  figure  it  lies  in  the  plane  of  the  figure  or  is  paral- 
lel to  it. 

(4)  With  a  surface,  it  is  parallel  to  some  line  or  element. 

160.  While  the  axis  of  rotation  may  be  made  to  assume  any  position 
to  the  planes  of  projection,  still  for  practical  purposes  it  ought  to  be  so 
placed  as  to  render  the  constructions  as  simple  as  possible.  Such  a  posi- 
tion is  one  in  which  the  axis  is  assumed  to  be  perpendicular  to  either 
plane  of  projection. 

1.     ROTATION   OF  THE   POINT. 

161.  In  order  to  effect  the  rotation  of  a  point  around  an  axis,  a 
perpendicular  must  always  be  passed  through  the  point  to  the  axis, 
giving  the  centre  of  rotation.  This  perpendicular,  which  is  the  Hne  of 
the  radius,  must  be  turned  into  the  required  position  and  the  original 
distance  of  the  point  from  the  axis  set  off  upon  it,  the  measurement 
being  made  from  the  foot  of  the  perpendicular  or  centre. 

162.  Problem. —  To  revolve  a  point  into  cither  plane  of  projection  around 
an  axis  lying  in  that  plane. 

Let  {x'y,   x'/)   be   the  given   axis  (Fig.   152),    and  {a\  a")   the   given 


*a" 


point.  The  perpendicular  draw^n  from  the  point  in  space  to  the  axis  is 
its  projecting  hne  to  H,  the  plane  m  which  that  axis  lies  ;  hence,  a"  is 
the  centre   of  rotation,  and  aa'  the  true   length   of  the  radius.     Through 


KOTATIOX   OF    THE   POINT. 


6l 


a"  draw  an  indefinite  perpendicular  to  the  axis,  and  lay  off  a  distance 
a"a^'  equal  to  aa'\  {a^",  a^)  are  the  projections  of  the  point  in  the  revolved 
position.     In  Fig-.   153  the  axis  is  a  line  of  the  vertical  plane. 

163.  PKt)l{LE.M.  —  7t?  revolve  a  point  into  either  plane  of  projection  around 
an  axis  lying  in  that  plane  tchen  neither  projection  of  the  point  eoineides  with, 
the  projection  of  the  axis. 

It  will   he   seen   from   Fig.   154  that  the  perpendicular  ac"  let  fall  from 


X' 


Fig.  105 


f 


/     y  /c""'^--. 


y'  «■ 


>a. 


the  point  a  to  the  axis  x")'"  is  the  hypothenuse  of  a  triangle  ac"a"y 
right-angled  at  a'',  in  which  aa" ,  the  perpendicular,  is  equal  to  the  dis- 
tance of  the  point  from  //,  and  a"c' ,  the  base,  is  equal  to  the  distance  of 
a"  from  the  axis. 

Hence,  in  Fig.  155.  {a,  a")  being  the  given  point  and  {x")'",  x'y')  the 
axis,  let  fall  from  a"  an  indefinite  perpendicular  to  x"y" ,  and  mark  the 
centre  of  rotation  {c" ,  c');    then  a"c"  is  the   base  of  a  triangle  of  which 


a'- 

ct\ 

^^^'' 

'V                 ,^  ■ 

1       ^"Sl-^       ! 

1    ^"Y     I 

!/"        ! 

-"     ~i 

0" 

6 


Kig 

x' 

1- 

jL^n' 

1 
1 

c 

i 

1 

y' 

1 
I 

x'^r 

' 

1 

x-'C'^v 

<■''     V n  ^ 

1 

1^" 

6            ^x     ! 

'^.. 

"n 

a" 

^a; 

aa',  the  height  above  //,  is  the  perpendicular.  At  a"  erect  a  perpen- 
dicular to  a"c",  and  lay  off  a  distance  a"a,"  equal  to  aa',  then  with  c" 
AS  a  centre  and  a  radius  equal  to  c"a/'  describe  an  arc  until  it  intersects 
a"c"  prolonged  ;   (<r,",  ^7/)  is  the  revolved  point  sought. 

164.  Problem. —  To  revolve  a  point  through  a  given  angle  around  a  hori- 
zontal axis. 

Let  {a\  a")  be  the  given  point,  and  {x'y',  x"y")  the  axis  (Fig.  156),  and 


62 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


let  8  be  the  given  angle.  As  the  plane  of  rotation  is  parallel  to  F,  the- 
radius  drawn  from  the  point  to  the  axis  is  likewise  parallel  to  V  during- 
its  entire  rotation  ;  hence,  a'c'  measures  the  true  length  of  the  radius,  and 
a"c"  is  its  horizontal  projection.  Draw  c'a^,  the  revolved  position  of  the 
radius,  so  that  a'c'a^  measures  the  required  angle ;  («/,  a/')  is  the  revolved 
point  sought. 

165.  Problem, —  To  revolve  a  point  tJiroiigJi  a  given  angle  around  a  vertical 
axis. 

Let  {a! ,  a")  be  the  given  point,  {x'y',  x"y")  the  axis,  and  d  the  angle 

(Fig.   157). 

As  the  plane  of  rotation  is  parallel  to  H,  the  radius  drawn  from  the 
point  to  the  axis  is  likewise  parallel  to  H  during  its  entire  rotation; 
hence,  a"c"  measures  the  true  length  of  the  radius,  and  a'c'  is  its  vertical 
projection.  Draw  c"a^",  the  revolved  position  of  the  radius,  so  that 
a"c"a^"  measures  the  required  angle ;  {a^", «/)  is  the  revolved  point  sought. 

166.  Problem. —  To  revolve  a  point  through  a  given  angle  and  around  any 
axis  parallel  to  the  horizontal  plane  of  projection. 

Let  {a,  a")  be  the  given  point,  {x'y',  x"y")  the  axis,  and  d  the  angle 
(Fig.   158). 

The  plane  of  the  circle  of  rotation,  being  at  right  angles  with  the  axis^ 
is  parallel  to  neither  coordijiate  plane;   hence  the  true  size  of  the  triangle,. 


m'       P'\  c'   \l>   a'l 


/  6 


(^ik 


the  hypothenuse  of  which  determines  the  length  of  the  radius,  must  be 
found.  This  is  effected  by  bringing  the  plane  of  rotation  parallel  to 
either  coordinate  plane ;    in  this  case  preferably  to  H. 

The  perpendicular  let  fall  from  the  given  point  (a',  a")  to  the  axis 
lies  in  a  plane  perpendicular  to  it,  giving  a  horizontal  trace  {a"c")  per- 
pendicular to  x"y"  and  cutting  the  axis  in  the  point  (c",  c').  The  dis- 
tance of  this  point  from  the  given  point  is  the  radius  of  rotation,  being 
the  hypothenuse  of  the  right-angled  triangle  whose  base  is  equal  to  c"l?" 
and  whose  perpendicular  is  equal  to  the  distance  ^7/  of  the  given  point 
above  the  horizontal  plane  of  the  axis. 

Turn  this  triangle  around  its  base  {l?"c",  b'c')  until  it  is  brought  paral- 


ROTATION  OF    THE   LINE.  63 

lei  to  H\  the  base  remains  fixed  while  the  given  point  is  projected  a* 
(rt,",  ^/),  the  distance  a"a^'  being  equal  to  a'b' ;  hence  a^'c"  is  the  hy- 
pothenuse  or  true  length  of  the  radius  let  fall  from  the  point  a  upon 
the  axis  xy  in  space.  From  c"  as  a  centre  and  with  this  radius  describe 
an  arc  whose  measure  is  the  given  angle  d  and  determine  a^' . 

By  a  counter-rotation  restore  the  plane  of  the  triangle  to  its  original 
position,  carrying  with  it  the  line  c"a^\  the  hypothenuse  of  a  seccjnd  tri- 
angle whose  base,  c"a^' ,  determines  «r/',  the  horizontal  projection  of  the 
revolved  point,  and  whose  perpendicular,  oi^"a^\  measured  above  fi^ 
determines  a,',  the  vertical  projection  of  that  point. 

167.  Problem. —  To  revolve  a  point  through  a  given  angle  around  any 
axis  parallel  to  the  vertical  plane  of  projection. 

Fig.  159  shows  constructions  which  are  similar  in  every  respect  to 
those  of  the  preceding  case. 

168.  Problem. —  To  revolve  a  point  through  a  given  angle  around  any  axis. 
Assume  a  new  ground-line  parallel  to  the  horizontal  projection  of  the 

axis,  reducing  the  case  to  that  of  Art.  166.  The  rotation  having  been 
effected  on  the  new  vertical  plane,  the  measurements  may  be  readily 
transferred  to  the  primitive  planes  of  projection. 

n.     ROTATION   OF  THE   LINE. 

169.  The  right  line  may  assume  one  of  three  distinct  positions  to  the 
axis  of  rotation : 

(ij  It  may  be  parallel  to  it,  describing  in  its  revolution  the  cylindri- 
cal surface. 

(2)  It  ma\'  incline  to  it — the  two  Iving  in  a  common  plane — describ- 
ing in  its  revolution  the  cone  of  one  or  two  nappes. 

(3)  It  may  have  any  position  not  in  the  plane  of  the  axis,  describing 
in   its  revolution  the  hyperboloid  of  one  nappe. 

If  the  revolving  line  be  a  curve,  it  will  generate  a  double-curved  surface. 

170.  Problenl — To  revolve  a  right  line  around  a  vertical  axis  and  through 
a  given  angle. 

Let  {a'b\  a"b")  be  the  given  line,  ix'y',  x"y")  the  axis,  and  6  the  angle 
(Fig.   160). 

During  the  rotation  of  the  line  each  point  thereof  describes  a  circle 
parallel  to  //.  the  relative  positions  of  line  and  axis  remaining  unaltered; 
hence  the  horizontal  projection  in  the  revolved  position  preserves  the 
same  distance  from  the  foot  x"  of  the  axis.  Draw  then  from  x"  a  per- 
pendicular x"o"  to  a"b",  thus  marking  the  shortest  distance  of  the  given 
line,  and  determine  the  vertical  projection  0'.  With  x"  as  a  centre  and 
x"o"  as  a  radius  describe  an  arc  whose  measure  is  the  given  angle  6, 
and  draw  the  new  horizontal  projection  o,"b^"  perpendicular  to  x"o^", 
the  second  point  b,"  being  obtained  by  measuring  from  o^"  a  distance 
equal  to  o"b". 

As  each  point  during  the  entire  rotation  preserves  a  fixed  distance 
from  //,  parallels  to  GL  drawn  through  o'  and  b'  determine  c,'  and  b^ 
through  which  the  vertical  projection  of  the  revolved  line  passes. 


64  ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 

In  Fig.  i6i   the  axis  has  been  assumed  to  be  perpendicular  to  V. 


/!> 


171.  Problem. —  To  revolve  a  right  line  until  it  becomes  parallel  to  either 
plane  of  projection. 

Let  {a'b',a"b")  be  the  given  line,  and  [x'y',  x"y")  the  axis, -and  let  it 
be  required  to  bring  the  line  parallel  to  V  (Fig.  162).  The  horizontal 
projection  of  the  line  will  be  parallel  to  GL\  hence  but  one  point  will 
be  required  to  determine  it. 


■"'  \ 

&; 

1  \ 

Va: 

05    i      \ 

^&r^^ 

o'{ 

S 

-^ 

a>^ 

a\ 

In  Fig.  163  the  line  has  been  brought  parallel  to  H,  the  axis  being 
perpendicular  to  V. 

172.  Problem. —  To  revolve  a  line  parallel  to  the  ground-line. 
Combine  the  two  solutions  of  Art.   171. 

173.  Problem. —  To  revolve  a  line  until  it  becomes  pcrpendictilar  to  the 
vertical  plane. 

Revolve  the  line  until  parallel  to  H,  then  turn  it  about  an  axis  per- 
pendicular to  H  until  its  horizontal  projection  assumes  a  position  per- 
pendicular to  GL. 

174.  Problem. —  To  revolve  a  line  until  it  becomes  perpendicular  to  the 
horizontal  plane 


A'OTA'J/OX  OF    THE   PLANE.  65 

Revolve  the  line  until  parallel  t(j  V,  then  turn  it  about  an  axis  jjer 
pendicular  to  V  until  its  vertical  projection  assumes  a  position  perpen- 
fllcular   to  GL. 

In  the  two  prcceclin«r  cases  the  construction  will  be  simplified  by 
passing-  the  axis  through   a  point  of  the   line. 

III.     ROTATION   OF  THE   PLANE. 

175.  A  plane  is  revolved  when  two  of  its  lines,  a  line  and  a  point  or 
three  points  are  turned  about  an  axis. 

176.  When  the  axis  is  a  line  of  the  revolving  plane,  the  determina- 
tion of  a  single  point  in  the  new  position  will  be  sufificient,  inasmuch 
as,  the  axis  always  remaining  a  line  of  the  plane  whatever  the  position 
assumed,  a  point  and  a  line  afford  the  requisite  data  for  that  plane. 

177.  When  the  axis  is  parallel  to  the  revolving  plane  it  is  to  be  ob- 
served that — 

(i)  But  one  line  in  the  plane  is  both  parallel  to  the  axis  and  meas- 
ures the  shortest  distance  between  the  two. 

(2)  The  line  so  determined  describes,  during  the  rotation,  a  cylinder 
to   whicli  the  plane  is  constantly  tangent. 

(3)  The  trace   of  the   plane  is  tangent  to  the  base  of  the  cylinder. 

178.  When  the  axis  intersects  the  revolving  plane,  the  determination 
of  a  single  line  in  the  new  position  will  fix  the  position  of  the  plane, 
the  point  of  intersection  between  the  axis  and  the  plane  in  connection 
with  that  line  affording  the  necessary  data. 

In  this  position  it  is  to  be  observed  that — 

(i)  One  line  of  the  given  plane  is  cut  by  the  plane  which  passes 
through  the  axis  and  is  perpendicular  to  that   plane. 

(2)  .This  line  describes,  during  rotation,  a  cone  whose  vertex  is  the 
point  of  intersection  of  the  axis  with  the  given  plane. 

(3)  The  revolving  plane  is  constantly  tangent  to  this  cone,  and  its 
trace  is  tangent  to  the  base. 

179.  Problem. —  To  revolve  a  plane  around  a  vertieal  axis  through  a 
given  angle. 

Let  {a'b\a"b")  and  {c'd\c"d")  be  two  horizontals  of  the  given  plane 
(Fig.   1 64"),  {x'y,  x"y")  the  axis,  and  S  the  angle. 

The  relative  positions  of  the  horizontals  to  each  other  and  to  the 
axis  are  not  altered  during  the  rotation,  each  point  of  the  lines  describ- 
ing a  circle  parallel  to  //.  Draw,  then,  x"c"  and  turn  it  through  the 
given  angle,  bringing  a"  to  ^^,",  and  e"  to  e^",  and  determining  the  hori- 
zontal projections  of  the  revolved  lines. 

The  vertical  projections  coincide  with  those  of  the  original  position. 

Should  the  traces  of  the  plane  be  given  (Fig.  165),  two  horizontals 
mav  again  be  taken,  one  of  which,  by  preference,  is  the  horizontal  trace 
//J/,  which  will  remain  during  the  entire  rotation  a  line  of  the  hori- 
zontal plane.  Letting  fall  from  x"  a  perpendicular  to  MM,  the  point  of 
intersection  /"  describes,  during  the  rotation,  a  circle  to  which  the  trace 


66 


ELEMENTS   OF  DESCRIPTIVE   GEOMETRY. 


is  constantly  tangent.  Measure  the  angle  S  and  draw  //"J/,,  the  new 
position  of  the  horizontal  trace,  thus  determining  the  ground-point  o  of 
the  vertical  trace.     A  second   point    may  be    obtained    by  revolving  any 


Eig.164 


X 

a' 

?>' 

7)'/ 

d[ 

c 

f?' 

ci; 

C'l 

V' 

Fig 

165 

x' 

V            . 

,>N 

vMy 

/ 

\vM 

d'ji. 

c[ 

.Pl-V 

^/   '■ 

y' 

\dr\ 

M') 

^\  xW/, 


horizontal  {c"d",  c'd')  of  the  plane  and  finding  its  vertical  piercing-point 

l8o.    Problem. —  To   revolve  a  plane  aroiuid  an  axis  perpendicular  to   the 
vertical  plane. 

In  Figs.   i66  and   167  the  constructions  are  precisely  similar  to  those 


,-C 


c" 

ct 

X" 

d': 

a" 

6' 

Cl'l 

h'l 

y" 

of   the   two    preceding    cases,  but    reversed    on    the   respective   planes   of 
projection   in  accordance  with  the  changed  position  of  the  axis. 

IV.     RABATTEMENT. 

181.  As  it  is  desirable  to  avoid  constructions  which,  by  the  position 
of  the  object  in  space,  may  appear  confused,  and  as  it  is  also  frequently 
necessary  in  practice  to  determine  the  true  size,  shape  and  position  of 
such  an  object,  or  of  its  separate  parts,  this  end  is  attained  by  the 
method  of  rotation  termed  rabattcnicnt. 

By    way    of    illustration,   assume   the    revohnng  object  to   be  a  plane  \ 


RABA  TTEMENT.  67 

then,  should  this  plane  be  revolved  around  any  line  which  is  parallel 
to  either  coordinate  plane  until  it  is  parallel  als<j  to  that  plane,  or 
should  it  be  revolved  around  any  line  lyin^^  in  either  coordinate  plane 
until  it  coincides  with  that  plane,   the   rotation  is  by  rabattcvioit. 

By  eoiintcr-rotation  is  understood  the  restoration  of  the  plane  to  its 
oriij^inal  position. 

182.  Problem. — Rabattcmcnt  of  a  plane  around  a  vertical,  the  plane  being- 
perpendicular  to  the  vertical  plane. 

Let  {x'y\  x"y")  be  the  vertical  and  {a  ,  a")  a  jjoint  of  the  plane  (Fig. 
168)  in  its  orii^inal  position. 

The  radius  of  rotation  drawn  as  a  perpendicular  from  the  point  {a\a"^ 
to  the  vertical  or  axis  is  parallel  to  //,  and  therefore  projected  in  a"c" 
in  its  true  size.  After  rotation  a" c"  becomes  parallel  to  V,  is  projected 
on  it  in  its  true  size,  and  on  //  in  a  line  parallel  to  GL.  Hence,  mak- 
\\\<g  c'a;  equal  to  c"a"  and  drawing  the  ordinate,  {a/,  a/')  are  the  pro- 
jections of  the   point   by  rabattement. 


ITif-.  1C« 


Kig.  100 


'1 


c< 


^k 


I       \ 


\  > 


j_   !      2'  \    \      y 


£7,-— -j— *;— -2,"  ia" 

I 

The  point  so  found  and  the  fixed  vertical  are  data  sufficient  to 
determine  the  plane  in  its  new  position.  Should  the  vertical  coincide 
with  the  vertical  plane  (Fig.  169),  the  rabattenienf  will  be  around  the  ver- 
tical trace  of  the  plane  as  an  axis. 

Converselv,  it  the  rabatteuicnt  has  been  effected,  the  original  position 
of  the  object   may   be  ascertained    by  reversing  the  operations. 

Thus,  having  the  point  {ii^,  a/'),  by  letting  fall  from  this  poii^i 
(Fig.  16S)  a  perpendicular  to  the  vertical  {x'v',  -t'V"),  by  restoring  the 
jtlanu  of  the  radius  and  axis  to  its  original  position  perpendicular  to  F, 
and  by  laying  off  the  distance  e"a"  equal  to  e'a^',  (a\  a")  are  the  pro- 
jections of  the  point  after  the  counter-rotation   has  been  made. 

183.    Proulem. — Rabattenient  of  a  plane  around  one  of  its  horizontals. 

Let  (.r  J'',  x")'")  be  the  horizontal  and  (//'.  a")  a  point  of  the  plane  in 
its  original  position  (Fig.  170). 

TIk'  radius  of  rotation  drawn  as  a  perpendicular  from  the  j)oint  {a',  a") 
to  the  horizontal  or  axis  is  i>arallcl  to  / '.  and  therefore  projected  in  aY 
in   its   true  sjze.     After  rotation  a"e'   becomes  parallel  to  H,  is  projected 


68 


ELEMENTS   OF  DESCRIPTIVE   GEOMETRY. 


on  it  in  its  true  size,  and  on  V  in  a  line  parallel  to  GL.  Hence,  making 
c"a^'  equal  to  c' a  and  drawing  the  ordinate,  («/',  «/)  are  the  projections 
of  the  point  by  rabattcnicnt.  The  point  so  found  and  the  fixed  hori- 
zontal are  data  sufficient  to  determine  the  plane  in  its  new  position. 

Should   the    Horizontal    coincide    with   H  (Fig.   171),   the    rabattement 
will  be  around  the  horizontal  trace  of  the  plane  as  an  axis. 

Fig.  iro 


0 


tka'' 


x' 

c'\     a[ 

tf 

^a-v 


\[< 


Conversely,  if  the  rabattement  has  been  effected,  the  original  portion 
of  the  object  may  be  ascertained  by  reversing  the  operations. 

184.  When  the  plane  of  the  object  inclines  to  both  planes  of  projec- 
tion, the  construction  may  be  effected  either  by  bringing  the  plane 
parallel  to  one  of  the  planes  of  projection  or  by  a  change  of  ground- 
line. 

The  following  examples  will  serve  to  illustrate  the  variations  in 
solution  in  connection  with  the  method  of  counter-rotation. 

185.  Problem. —  Through  a  given  point  of  a  plane  to  draw  a  rectangle 
in  that  plane. 


RABA  TTEMENT. 


09 


Let  {VM,  HM)  be  the  plane,  and  a'  the  vertical  projection  of  the 
point  (Fig.  172). 

Determine  the  horizontal  projection  a"  of  the  point  by  means  of  a 
horizontal,  and  assume  a  new  vertical  plane  perpendicular  to  the  given 
plane ;  HN  is  its  horizontal  trace  or  ground-line  perpendicular  to  HM. 
Since  the  planes  M  and  N  are  at  right  angles,  the  new  vertical  projec- 
tion rt/,  found  by  making  «,<?/  equal  to  aa ,  is  a  point  in  the  line  ot 
intersection  or  trace  between  them;  the  second  point  being  ;///,  where 
the  horizontal  traces  cross  each  other. 

Revolve  the  plane  M  around  the  line  w,'^/  as  an  axis  into  the  new 
vertical  plane ;  find  a^  by  making  a^a^  equal  to  oc^a" ,  and  construct  the 
required  rectangle  a^b^c^d^. 

By  a  counter-rotation  bringing  the  plane  of  the  rectangle  back  to  its 
original  position,  b^d^  is  the  new  vertical  projection,  from  which  the 
primitive  horizontal  may  be  found  by  making  /i/"  equal  to  b^b^,  Yx"^' 
equal  to  ^,V/,  and  d^d"   equal  to  d^d^. 

186.    Problem. —  To  construct  a  triangle  on  any  line  of  a  given  plane. 

Let  (FJ/,  HM)  be  the  given  plane  (Fig.  173),  and  a'b'  the  vertical 
projection  of  the  Hne. 

Determine  the  horizontal  projection  {a"b")  by  means  of  horizontals, 
and  assume  an  auxiliary  plane  S,  perpendicular  to  V  and  to  the  given 
plane   M\    VS  is  its  vertical  trace  or  ground-line.     The  intersection  of  .S 


and  M  passes  through  the  supplementary  projection  (^,7'/)  found  by 
making  «/?/  equal  to  aa" ,  and  (i.b^  equal  to  /?//'. 

By  rabattiinent  around  the  vertical  trace  J'M,  the  points  a  and  b 
describe  circles  parallel  to  5",  the  projections  of  which  are  perpendicular 
to  FJ/;  hence  a^b^'  is  the  revolved  projection  of  the  line  ujuin  which 
the  triangle  may  be  constructed. 

By  counter-rotation    r/  falls    in   supplementary   projection   at   c,',  from 


TO 


ELEMENTS   OF  DESCRIPTIVE   GEOMETRY. 


CHAPTER  VIII. 
DISTANCES  AND   PERPENDICULARS. 

I.      DISTANCE    OF    TWO    POINTS. 

188.  Problem. —  To  determine  the  distanee  between  tzvo  points  given  by 
their  projeetions. 

The  distance  between  two  points  is  measured  by  the  right  line  which 
joins  them. 

(i)  Let  {a',  a")  and  {b\  b")  be  the  given  points  (Fig.  174).  Bring  the 
plane,  horizontally  projecting  the  line  joining  the  two  points,  parallel  to 
V  by  rabatteinent  around  any  vertical,  preferably  that  passing  through  the 
point  {a' ,  a").  This  point  remains  fixed  during  the  rotation,  and  {b' ,  b") 
assumes  the  position  (/?/,  /;/');   a'b^  is  therefore  the  distance  required. 

Fig.  17-4 

h  r!  ITitr.  ITS 


-.4 


A- 

-,X 

^-' 

_,--r 

^''  ^ 

Y' 

™"-k- 

lli' 

a\ 

"j 

■    ^t?, 

a.\. 


V6 


.H--- 


h". 


(2)  Bring  the  plane,  vertically  projecting  the  line  which  joins  the  two 
points,  parallel  to  H  by  rabattement  around  any  horizontal,  preferably 
that  passing  through  point  {a\  a")  (Fig.  175).  This  point  remains  fixed 
during,  the  rotation,  and  (//,  b")  assumes  the  position  (/;/,  b^') ;  a"b^'  is 
therefore  the  distance  required. 

(3)  Bring  the  plane  (Fig.   176),  horizontally  projecting  the  line  which 


av:--'--- k-,K 

¥1 

/ 1 
/  1 

/ 

■  /    1 

Kig.  i.rr 

1      1 

t"* 

1 

1 

1                                         1              1 

-Jf«" 

■joins    the    points,    parallel   to    H  by    rabattemen 


around    any    horizontal, 


preferably  that  passing  through  («',  a").     This   point   remains   fixed,  and 


DISTANCES  AND  PERPENDICULARS. 


71 


the  point  {b\  b")  assumes  the  position  {b,',  b^')  by  making  the  perpen- 
dicular b"b^'  equal  to  c'b' ;   hence  a"b^'  is  the  distance  required. 

(4)  Bring  the  plane  (Fig.  177),  vertically  projecting  the  line  which 
joins  the  two  points,  parallel  to  V  by  rabattcmcnt  around  any  vertical, 
preferably  that  passing  through  {a\  a").  This  point  remains  fixed,  and 
the  point  {b' ,  b")  assumes  the  position  {b^',  b^")  by  making  the  perpen- 
dicular b'b^'  equal  to  c"'b" ;    hence  a'b,'  is  the  distance  required. 

189.  Problem. —  Upon  a  given  line  to  measure  a  given  distance  from  either 
extremity. 

Let  (<?',  a")  be  the  extremity  from  which  the  measurement  is  to  be 
made  (Figs.   178,   179),  and  {b\  b")  any  other  point  of  the  given  line. 

Vie.  178      ,.^  ,/ 


I-V^— 4— lb, 


Kig.  !•: 

1'- 


l-'V-'-n 


a"^. 


h 

four 


methods    parallel    to 


Bring  the  line  by  any  of  the  precedim 
either  coordinate  plane,  and  measure  upon  the  projection  so  determined 
the  required  length.  By  a  counter-rotation  restore  the  dividing  point 
(<",',  r,")  to  the   primitive  projections;   {ac\a"c")  is  the  distance  sought. 

ir.      DISTANXE    OF   POINT    FROM    LINE. 

190.  Problem. —  To  determine  the  perpendicular  betiveen  a  point  and  a 
line  given   by  their  projections. 

The  point  and  line  fixing  the  position  of  a  plane,  their  distance  from 
each  other  may  be  found  by  the  rabattement  of  that  plane. 

i^is.  ISO  y  ^-^X"-—-^  "1 


1    y     17-^ 

i^ 

''1^^ 

\ 

jjf 

\ 
\ 

V 

"C'l 

(i)  Let  {a\  a")  be  the  given  point,  and  (b'c',  b"c")  the  given  line  (Fig.  i8o> 


72 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


Bring  the  plane  of  these  two  by  rabattcnicnt  around  a  horizontal, 
preferably  that  which  passes  through  the  point  {a',  a").  During  rota- 
tion this  point  remains  fixed,  and  the  line  be  assumes  the  position  {b,"c^", 
b,'c,')  (Art.  183);  hence,  lettmg  fall  a  perpendicular  {a"o,")  upon  b,"c^", 
a"o,"  is  the  horizontal  projection  of  the  perpendicular  sought. 

By  counter-rotation  the  foot  of  the  perpendicular  (^/,  o^")  falls  in  {o\ 
o"),  and  hence  ia'o' ,  a"o")  are  the  primitive  projections  of  the  line  let  fall 
from  {a\  a")  upon  the  given  line. 

In  Fig.  181  the  rabattenient  has  been  effected  around  a  vertical  {a'c', 
a"c")  passing  through  the  given  point  {a',  a"). 

During  rotation  this  point  remains  fixed,  as  does  likewise  {c',  c")  of 
the  given  line ;  hence  b^'c'  will  be  the  line  revolved,  and  a'o^'  the  perpen- 
dicular sought. 

By  counter-rotation  {a'o',  a"o")  are  the  projections  of  the  perpen- 
dicular  let  fall  from  the  given  point  upon  the  line. 

2d  Solution. — The  problem  may  also  be  solved  by  passing  through 
the  given  point  a  plane  perpendicular  to  the  given  line,  when  the  line 
joining  this  point  with  the  point  of  intersection  between  the  given  line 
and  auxiliary  plane  determines  the  required  perpendicular. 

Thus  in  Fig.  182  let  {a',  a")  be  the  given  point,  and  {b'c,  b"c")  the 
given  line. 

To  fix  the  position  of  the  auxiliary  plane  three  steps  are  necessary : 


IPig.lSS 


ris^xss 


to  pass  any  plane  through  the  given  line,  to  let  fall  upon  this  plane  a 
perpendicular  from  the  given  point,  and  to  draw  from  the  foot  of  the 
perpendicular  a  second  perpendicular  to  the  given  line.  The  two  per- 
pendiculars determine  the  required   plane. 

Hence,  assume  the  plane  horizontally  projecting  the  line  in  b"c"  as 
the  plane  of  the  line;  the  perpendicular  let  fall  upon  it  from  a  in  space 
gives  the  projections  {a"d",  ad').     From   the  foot  id',  d")  of  this  perpen- 


DISTANCES  AND  PERPENDICULARS. 


71 


dicular  draw  a  perpendicular  to  {b'c',  b"c"),  which  may  be  effected  by 
the  rabattcmcnt  of  the  j)rojectincr  plane  of  the  line  around  one  of  its 
horizontals,  preferably  that  which  passes  through  the  point  {d' ,  d"). 
During  rotation  the  horizontal  remains  fixed,  and  the  line  be  assumes 
the  position  {b^'c^',  b^c^)\  hence  {d"c^',  d't\)  is  the  second  perpendicular 
sought. 

By  counter-rotation  (r,",  c^)  falls  at  (r",  c')\  then  {a"d'\  a'd')  and 
(d"e",  d'e')  are  the  two  lines  which  determine  the  position  of  the  re- 
quired  plane,  and  {a'c',  a"c")  is  the  perpendicular  drawn  from  tlie  given 
point  to  the  given  line. 

The  distance  of  the  i)()int  {a\  a")  from  the  given  line  is  equal  to  the 
hypothenuse  of  a  right-angled  triangle,  of  which  the  base  is  the  distance 
a"d"  of  the  point  from  the  plane  of  the  line,  and  the  perpendicular  the 
distance  d"€^'  of  its  foot  from  the  given  line;  hence  drawing  d"a  per- 
pendicular to  d"e^'  and  equal  to  d"a" ,  ae,"  is  the  distance  of  the  point 
from  the  line. 

In  Fig.  183  the  plane  projecting  the  line  on  V  in  b'c'  being  assumed 
as  the  plane  of  the  line,  the  operations  are  a  repetition  of  those  of 
the  preceding  case. 

191.  Problem. — From  a  given  point  as  a  vertex,  to  construet  a  triangle 
on  a  given  line. 

Let  {a',  a")  be  the  given  point  (Fig.  184),  and  {b'c',  b"c")  the  line. 


Bring  the  plane  of  these  two  by  rabattenient  around  a  horizontal  {a'o\ 
a"o",  by  the  operations  of  the  preceding  problems.  From  a"  as  a  centre 
and  with  radii  equal  to  the  length  of  the  given  sides  of  the  triangle, 
describe  arcs  intersecting  ^/V,"  in  </,"  and  e^" ;  a"d,"e^"  is  the  triangle 
in  its  true  size. 

By  counter-rotation  the  points  of  the  triangle  are  projected  at  (/',  e') 
and  {d",  d')\  hence  {a"d"e",  a'd'c')  are  the  projections  of  the  required 
triangle  on  the  primitive  planes. 


74 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


In  Fig.   185   the    rahattcmcnt   has    been  effected  around  a  vertical  {a'o\ 
a"o"')\   hence  b^c^  is  the  revolved   line  to   which  the  measurements  from 


Fi'g,.  iss 


the    given    point   («',   «")    are   to   be   made.      The   triangle   a'd^ 

resented   in    its   true   size,  and  {a'd'e',  a"d"e")   are   its   projections  on   the 

primitive  planes  after  the  counter-rotation. 


III.      DISTANCE    OF    POINT    FROM    PLANE. 

192.  The  distance  of  a  point  from  a  plane  is  measured  by  the  per- 
pendicular let  fall  from  the  point  upon  that  plane.  This  perpendicular 
ma}^  be  found  by  passing  through  the  point  any  auxiliary  plane  perpen- 
dicular to  the  given  plane,  by  determining  the  line  of  intersection  be- 
tween the  two  planes,  and  by  drawing  to  this  line  a  perpendicular  from 
the  point.     The  case  resolves  itself,  therefore,  into  that  of  Art.   190. 

193.  Problem. —  To  determine  the  distance  between  a  point  and  a  plane. 
The   following  are  various   solutions   showing  the  application  of  pre- 
ceding principles. 

(i)  Let(«',  a")  be  the  given  point,  {b'c\  b"c")  a  horizontal,  and  {d' ,  d") 
a  point  of  the  given  plane  (Fig.   186). 

Through  the  point  {a',  a")  pass  a  plane  perpendicular  to  the  hori- 
zontal ib'c',  b"c") ;  its  horizontal  trace  by  Art.  104  passes  through  a" 
and  is  perpendicular  to  b"e"  at  c'\  which  is  a  point  of  the  line  of  in- 
tersection between  this  auxiliary  plane  and  the  given  plane.  The  hori- 
zontal {d'e',  d"e")  determines  the  second  point  {c\  e"),  whence  {c'e',  c"e") 
is  the  line  of  intersection  sought. 

By  the  rabattement  of  the  auxiliary  plane — which  contains  the  given 
point  and  the  line  of  intersection — around  a  horizontal,  preferably  that 
which  passes  through  the  point  {c',  c"),  the  points  {e',  e")  and  {a\  a") 
fall  at  {e,",  e,')  and  (a/\  a/)  respectively;  hence  a/'o/'  is  the  rabattement 
of  the  perpendicular  drawn   from  the  point  {a! ,  a")   to  the  given  plane. 


DISTANCES  AND   PERPENDICULARS. 


75 


By    counter-rotation    o^'    is    projected    in   o'\   and    {a"o'\  a'o')   are    the 
(rimitive   projections  of  the   required   distance. 

(2)    Let   (FJ/,  MM)    be    the    given    plane,    and  {a',  a")   the    point    (Fig. 
1S7). 

Let    fall    from    the    point   a   perpendicular   upon    the    plane  (Art.    103), 
y     and    find    the    piercing-point    thereon  ;     the    distance    between     the    two 


ISO      ^ 

l> 

_i 

'1  1     1 

6-'/ 1|  1        e[    a\ 

1                  1  II  1 

^> 

Ti?ie.l87 


points  measures  the  distance  from  the  plane.  To  determine  this,  assume 
a  new  vertical  plane,  of  which  C.Z,  is  the  ground-line  perpendicular  to 
the  given  plane,  and  hence  parallel  to  the  line  {a'b',  a''b")\  a^b^  is  the 
new  vertical  projection  marking  the  real  length  of  the  line,  and  vi'b^ 
the  new  vertical  trace  to   which   it  will  be  perpendicular. 

(3)    Let   (FJ/,   HM)   be   the    given    plane,  and   {a ,  a")  the    point  (Fig. 
188). 


T^ig.lSO 


Bring  the  plane  J/  perpendicular  to  V  by  rotation  around  a  vertical 
axis  passing  through  the  given  point  (a\  <j");  (//J/,.  TJ/,)  are  the  traces 
in  the  new  position  (Art.  179V  During  the  rotation  the  point  (<?',  a") 
remains  fixed,  and  the  line  {a'b',  (j"b")  drawn  perpendicular  to  J)/  be- 
comes parallel  to  [';  hence  a'b^'  drawn  perpendicular  to  f'J/j  is  the 
j-equircd  perpendicular  in  its  true   length. 


76  ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 

Restoring  the  plane  to  its  original  position,  {a'b\  a"b")  are  its  pro- 
jections in  the  original  position. 

(4)  Let  iyM,  HM)  be  the  given  plane,  and  {a',  a")  the  point  (Fig. 
189). 

From  {a',  a")  let  fall  a  perpendicular  upon  the  plane,  find  its  piercing- 
point  thereon,  and  revolve  the  line  {a'b' ,  a"b")  parallel  to  either  coor- 
dinate plane  (Art.   171). 

194.  Problem. —  Through  a  given  point  in  a  line  to  pass  a  plane  perpcn- 
diailar  to  that  line. 

The  given  point  is  a  point  of  the  required  plane ;  hence  a  horizontal 
or  vertical  passed  through  it  determines  by  its  piercing-point  the  re- 
quired traces. 

195.  Problem. —  Through  a  given  point  to  pass  a  plane  perpendicular  to 
a  given  plane. 

Through  the  point  let  fall  a  perpendicular  upon  the  plane.  Every 
plane  passing  through  this  line  will  be  in  the  required  position. 

196.  Problem. — To  determine  the  distance  between  two  parallel  planes. 
Resolved  into  the  problem  Art.   193. 

197.  Problem. —  To  erect  a  rigJit  prism  standing  upon  a  given  plane. 
Let  {HM,   VM)  be  the  given  plane,  and  a'b'c'   the  vertical   projection: 

of  the  base  of  the  prism  (Fig.  190).     Find  the  horizontal  projection  a"b"c'^ 
by   means  of   horizontals,  and  draw  indefinite  lines  perpendicular  to  the 


plane  for  the  projections  of  the  edges  in  position.  Rotate  any  plane,  J 
projecting  an  edge  on  F,  by  rabattement  around  its  vertical  trace,  as, 
for  example,  the  projecting  plane  of  the  edge  [ad)  in  space;  the  point 
«/  is  the  rabattement  of  a  of  the  base,  and  a,'o'  that  of  the  line  of  in- 
tersection between  the  plane  M  and  the  projecting  plane.  But  the  edge 
is   perpendicular   to   every  line   of   the   given   plane   passing   through   its 


I 


D/S7AXCES  AXD    PERPENDICULARS.  77 

foot;  hence  ^,V,'  drawn  perpendicular  to  o'a^  and  made  equal  to  the 
required  altitude  of  the  prism  is  the  rabattemcnt  of  the  edge  ad. 

By  counter-rotation  d^  falls  at  d' \  hence  {ad\  a"d")  are  the  projec- 
tions of  the  edge  on  the  primitive  planes.  Lay  off  from  {b',  c')  and 
{b",  ^"),' respectively,  distances  equal  to  a'd'  and  a"d",  and  complete  the 
projections  of  the  prism  by  joining  the  extremities  of  the  edges  so  found. 

198.  FK(niLV.y\.— Given  a  plane  and  the  rabattemcnt  of  the  base  of  a  right 
pyramid  standing  on  that  plane,  to  determine  the  projections  of  the  pyramid 
after  counter-rotation. 

Let  {VM,HM)  be  the  given  plane  (Fig.  191),  and  a.,"b,"c.^'d^"  the  ra- 
battemcnt of  the  base  around  the  horizontal  trace  HM. 

The  ordinates  aa^"fib„y,  etc.,  measure  the  perpendicular  distances  of 
the  points  of  the  base  from  the  trace  HM  both  before  and  after  rota- 
tion ;  hence,  if  the  plane  be  counter-rotated,  these  distances  will  be  the 
hvpothenuses  of  right-angled  triangles,  of  which  the  perpendiculars  are 
the  heights  of  the  points  above  H,  and  the  bases  the  distances  of  their 
horizontal  projections  from  the  trace  HM. 

Assume  a  new  vertical  plane  which  shall  pass  through  ee^'  the  middle 


Kig.  19i 


/  V  ,'.-7^-^ 


point  of  the  base  and  be  perpendicular  to  the  plane  M;  its  horizontal  trace 
r/'^'"  is  perpeiuiicular  to  HM.  By  rabattemcnt  around  its  horizontal  trace, 
the  line  of  intersection  with  the  given  plane  falls  at  ev/  and  contains 
the  new  vertical  projection  a^b'c.'d'  of  the  base.  To  determine  this 
projection  lay  off  upon  the  line  er/  the  distances  of  the  points  «/',  A/',  etc., 
from  HM.  At  e'  erect  the  given  altitude  f,'//.  ^"cl  complete  the  new- 
vertical  projection  of  the  pyramid,  from  which  the  primitive  projections 
may  readily  be  found  by  the  usual  methods. 


78 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


IV.      SHORTEST    DISTANCE    BETWEEN    TWO     LINES. 

199.  A  line  may  be  a  common  perpendicular  to  two  right  lines  either 
in  direction  or  position. 

It  is  a  perpendicular  in  direction  when,  should  both  lines  be  moved 
parallel  to  their  original  positions  respectively,  and  towards  the  assumed 
line,  they  will  intersect  it  at  right  angles. 

Thus,  if  any  plane  [VM,  HM)  be  assumed  as  perpendicular  to  the 
first  line  {a' b\  a"b")  (Fig.  192),  then  any  line  parallel  to  that  plane  will 
be  a  perpendicular  in  direction  to  the  given  line ;  in  like  manner,  if  any 
plane  {VP,  HP)  be  assumed  as  perpendicular  to  the  second  line,  then 
any  line  parallel  to  that  plane  will  be  a  perpendicular  in  direction  to 
the  given  line.  Hence  the  direction  sought  is  parallel  to  both  planes^ 
or  to  their  line  of  intersection  {m'p',  m"p"). 

Fig.  193 


0  :bm\  \  1   /  ^  ^ 


The  perpendicular  in  direction  may  also  be  fixed  in  its  position  by 
determining  a  plane  parallel  to  the  two  given  lines,  when  any  perpen- 
dicular to  the  plane  so  found  is  the  direction  sought. 

Thus,  in  Fig.  193,  through  any  assumed  point  (/',  p")  lead  lines, 
parallel  to  the  given  lines,  and  find  a  vertical  and  a  horizontal  of  the 
plane  which  they  determine  (Art.  82);  the  perpendicular  (g'o',  g"o")  to 
this  plane  is  the  direction  sought. 

200.  The  common  perpendicular  in  position  to  two  right  lines  is  that 
line  which  intersects  both  at  right  angles. 

When  the  two  lines  are  parallel   they  lie   in   the   same   plane;    hence, 
any  Une  which   intersects   them    must   also   be   a   line   of  that   plane.      \n     j 
this  special  case  any  number  of  lines  may  be   perpendicular  in  position. 

When  the  lines  are  not  parallel  there  is  but  one  perpendicular  in. 
position.  As  the  perpendicular  in  direction  is  parallel  to  the  perpen- 
dicular  in  position,  planes  passing  through  the  two  lines  and  parallel,, 
respectively,  to  the  perpendicular  in  direction  must  intersect  in  the  per- 
pendicular  in  position.  Thus,  in  Fig.  194,  the  perpendicular  in  direction 
p'm',  p"m"  having   been   determined  as  in   Fig.   192,   through   each   of  the 


DISTANCES  AXD   PERPEXDICULARS. 


79 


lines  ab  and  cd  in  space  pass  a  plane  parallel  to  this  line  (Art.  94). 
The  line  of  intersection  between  the  two  planes  being  parallel  to 
{p'm\  P"}n"),  a   single    point   /",  determined   by  means  of  the  horizontals 


Kig.  104r 


rb  and  cf,  is  sufficient  to  determine  ^g'i',  g"^"),  the  position  of  the 
common  perpendicular  sought.  The  perpendicular  ~  in  position  having 
been  found,  the  points  in  which  it  intersects  the  given  lines  are  the 
extremities  of  the  line  which  measures  the  shortest  distance  between 
them. 

201.  Problem. —  To  determine  the  line  vieasuring  the  shortest  distance 
betzveen  tn'o  right  lines   not  in  the  same  plane. 

Let  ab  and  ed  be  the  given   lines  (Fig.   195). 

(i)  Pass  through  the  line  ab  a  plane  parallel  to  cd \  HM  and  I'M 
are  its  traces. 

(2)  From  any  point  of  cd,  as  p,  let  fall  a  perpendicular  up«^n  the 
plane  M,  and   find   its  piercing-point  (w',  w")  thereon. 

(3)  Through  (;//',  m")  draw  a  parallel  to  {c'd\  c"d")\  then  are  {ma'. 
m"a")  the  projections  of  cd  on  the  plane  M \  hence,  ma  and  ab  lying  in 
the  same  plane  intersect  each  other  in  the  point  {a\  a"). 

(4)  From  the  point  {a',  a")  erect  a  perpendicular  to  the  plane  J/, 
which  is  the  i)crpendicular  in  position  sought.  For  (a'd\  a"d")  is  per- 
y)cndicular  to  {a  b\  a"b")  and  {a'm',  a"m"),  since  they  pass  through  its 
foot  {ii\  a")  and  lie  in  the  plane  J/,  to  which  it  is  perpendicular:  hence 
it  is  a  projecting  line  of  the  plane  projecting  cd  upon  J/,  and  must  in- 
tersect cd  in   the  point  d. 

The  true  length  of  the  line  {a'd',  a"d")  may  be  determined  bv  Art. 
193- 


8o 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


202.    When    one    of    the    lines    is    perpendicular    to    either    coordinate 
plane  the  construction  is  greatly  simplified. 


Fig.  193 


Thus  in  Fig.  196  the  line  ab  is  vertical;  hence  the  plane  passed 
through  cd  parallel  to  it  is  also  vertical,  and  the  perpendicular  in  posi- 
tion horizontal.  The  plane  passed  through  ab  parallel  to  the  line  of 
shortest  distance  gives  a  horizontal  trace  a"e"  perpendicular  to  c"d" ; 
hence  {a'e\  a" c")  are  the  projections  of  the  line  of  shortest  distance: 
and  since  this  line  is  horizontal,  a"e"  is  its  true  measurement. 


Fis.  196 


In  Fig.  197,  as  the  plane  passed  through  cd  parallel  to  ab  is  perpea 
dicular  to   V,  the  perpendicular  in   position   is  parallel  to  V. 

The  plane  passed  through  ab  parallel  to  the  line  of  shortest  distance 
gives  a  vertical  trace  a'e'  perpendicular  to  c'd' ;  hence  {a'e' ,  a"e")  are  the 
projections  of  the  shortest  distance  ;  and  since  the  line  is  parallel  to  V, 
{a'e')  is  its  true  measurement. 


DISTANCES  AND  PERPENDICULARS. 


8i 


203.  Should  both  lines  be  parallel  to  cither  coordinate  plane  the 
construction  will  be  still  further  simplitied. 

Thus,  in  Fig.  198,  the  lines  ab  and  cd  being  parallel  to  //,  the  plane 
passed  through  ab  is  horizontal,  and  the  perpendicular  in  direction  is 
vertical ;  hence  the  planes  passing  through  the  given  lines  and  parallel 
to  this  perpendicular  are  also  vertical,  and  give  c'f\  the  vertical  projec, 
tion  of  the  line  of  shortest  distance,   in   its  true   measurement. 


r' 

e' 

d' 

a' 

\r 

6' 

C"  r'  ,/ 

a 

In  Fig.  199  similar  reasoning  will  show  (//',  e"f")  to  be  the  projec 
>,. 'US  of  the  line  of  shortest  distance,  and  c"f"  its  true   measurement. 

204.  When  the  two  lines  lie  in  the  same  plane  they  must  either  in. 
tc'scct   or  be   parallel. 

In  the  first  case  the  line  of  shortest  distance  passes  through  the 
p(/iiit  of  intersection,  and  hence  is  reduced  to  a  point. 

In  the  second  case  the  plane  passing  through  one  line  parallel  to 
the  other  will  be  indeterminate  and  the  application  of  the  general  rule 
will  fail.  But  the  shortest  distance  may  readily  be  found  by  letting  fall 
irom  any  point  of  either  line  a  perpendicular  to  the  other  (Art.   190). 

205.  Problem. — Given  the  horizontal  projections  of  tivo  lines  and  the  two 
projections  of  their  line  of  shortest  distance,  to  determine  the  vertical  projections. 

,  Fis-SOO 


Let  a"b"  and  c"d"  (Fig.  200)  be  the  horizontal  projections  of  the  two 
lines,  and  (//',  e"f")  the  projections  of  the  line  of  shortest  distance. 


82 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


The  line  ab  is  perpendicular  to  cf,  and  hence  lies  in  a  plane  passing 
through  e  and  perpendicular  to  cf.  Determine  a  horizontal  [e'g',  e"g") 
and  a  vertical  {e'h',  c" li")  of  this  plane,  and  assume  any  third  line  of 
this  same  plane,  as,  k)r  example,  the  one  whose  horizontal  projection  is 
a"h";   find  its  vertical  projection  h'g',  and  a'   is   projected  in  it. 

By  a  similar  process  of  reasoning  the  vertical  projection  c'd'  may  be 
determined. 

206.  Problem. — Given  the  projections  of  a  line  and  the  horizontal  pro- 
jections of  a  second  line  and  the  line  of  shortest  distance,  to  determine  their 
vertical  projections. 

Let  ia'b',  a"b")  be  the  given  line  (Fig.  201),  and    /'/",  c"d"  the  hori- 


Kig.  soa 


zontal  projections  of  the  shortest  distance  and  the  second  line  respect- 
ively. 

The  line  r/,  being  perpendicular  to  ab,  hes  in  a  plane  passing  through 
e  and  perpendicular  to  ab.  Determine  a  horizontal  {e"g" ,  e'g')  and  a 
vertical  [e'h',  e"h")  of  this  plane,  assume  any  third  line  of  the  plane,  as 
that  whose  horizontal  projection  is  f'h",  find  //'  and  g' ,  and  /'  hes  in 
the  line  which   connects  them. 

The  vertical  projection  c'd'  of  the  second  line  may  be  found  as  in 
the  preceding  case. 


ANGLES. 


83 


CHAPTER    IX. 

ANGLES. 


I.      ANGLES    BETWEEN     RRIHT     LINES. 

207.  Ris^ht  lines  are  either  parallel  or  intersect,  or  without  being 
parallel  do  not  intersect.  In  every  case  the  plane  angle  obtained  by 
passing  through  a  point  lines  parallel  to  the  lines  in  space  is  termed 
the  angh'  of  the  lines. 

To  determine  these  angles,  therefore,  it   is  simply  necessary  to  bring 

the  plane  of  the  angle  parallel  to  either  coordinate  plane. 

208.  Proble>L — To  determine  the  angle  beticeeti  tico  lines  iL'hose  projec- 
tions are  given. 

The  lines  mav  lie  in  the  same  plane,  whicii  will  be  indicated  when 
the  points  of  intersection  of  the  projections  lie  in  a  common  perpen- 
dicular to  GL  (Art.  53). 

Let  {a'b\  a"b")  and  {c'd',  c"(i")  be  the  given  lines  (Fig.  202).  and  {0' , 
o")  their  point  of  intersection. 

Find  the  horizontal  piercing-points  a"  and  il",  through  which  passes 
the  horizontal  trace  of  the  plane  of  the  lines.  Turn  this  plane  into  // 
by  rotating  (</,  o")  around  a"d"  as  an  axis  (Art.  163),  when  a  is  the  re- 
quired angle. 

x> 


In  Fig.  203  the  piercing-points  are  not  immediately  available.  The 
'solution  in  this  case  is  effected  by  the  rahattement  of  the  {)lane  of  the 
angle  around  anyi  horizontal,  as  ya'd\  a"d"). 

Should  one  of  the  intersecting  lines  ah  be  horizontal  ( P'ig.  204),  the 
horizcMital  trace  e"/"  of  the  plane  of  the  angle  will  be  parallel  to  a"b" 
(Art.  /L).  Around  this  trace  as  an  axis  revolve  the  point  {o' ,  o"),  and  a 
is  the  anole  souirht. 


84 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


If  the  point  of  intersection  lies  in  H  (Fig.  205),  through  any  point 
{e\  e")  of  one  hne  lead  a  line  {e'g' ,  e"g")  parallel  to  the  other ;  the 
angle  between  these  two,  which  is  equal  to  the  angle  sought,  may  be 
determined  as  in  Fig.  202. 


Fig,  S05 


209.  Problem. —  To  determine  the  angles  wJiicJi  a  line  in  space  makes 
with  its  projections. 

Bring,  the  plane,  projecting  the  line  horizontali}-,  by  rabatteinent 
around  a"b''  (Fig.  206),  and  as  the  line  ab  in  space  and  its  projection 
a"b"  lie  in  a  common  plane,  their  prolongations  intersect  in  a  point 
{a' ,  a")  and  include  the  angle  a,  or  the  angle  of  inclination  to  the  hori- 
zontal plane  of  projection. 

The   rabattement  measures  the  true  length  of  the  line  in  space. 

A   similar  construction  determines  the  angle  ^  for  the  vertical  plane. 

In  Fig.  207  the  projecting  plane  of  the  line  is  brought  by  rabatte- 
ment around   either   a  horizontal  {a"c",  a'c')  in  order   to  determine   oc,   or 

Fig.  sor 


Fig.  206 

-4'   1 

i    i 

A^ 

o:"^-^ 


-^^: 


around  a  vertical  {b'd',  b"d")  to  determine  /?. 

210.  Problem. —  To  determine  the  angle  betzveen  the  traces  of  a  given 
'^lane. 

The  sides  of  the  angle  or  the  traces  being  respectively  a  vertical  and 
a  horizontal  of  the  plane  (Fig.  208),  the  solution  is  effected  as  in  Art. 
-08. 

211,  Proble]\l — The  projections  of  two  iiitersecting  lines  being  given,  to 
bisect  or  otherwise  divide  their  angles. 


ANGLES. 


85 


Let  {a'b\  a"h")  and  {c'd',  c"d")  be  the  ^iven  lines  {Y\^.  209). 

Determine  the  angle  ot  between  them,  as  in  the  preceding  cases,  and 
draw  the  lines  o^'p"  and  o^'q"  which  divide  the  angles  in  tlie  required 
ratios.     As   these   lines  lie   in   the   plane  of    the   angle,  their   feet  (/",  p'^ 


ITie.SOB 


Fig.aoo 


and  {(/",  q')  must  be  contained  in  the  horizontal  trace  a"d"  of  that  plane; 
hence,  during  counter-rotation,  they  remain  fixed  in  position,  and  the 
projections  of  the  bisecting  or  other  dividing  lines  sought  must  pass 
through  the  points  (/',  p")  {q',  q")  and  the  vertex  {o',  0")  of  the  angle. 
-^  •212.  Problem. —  Through  a  ^ivcn  point  to  pass  a  line  ivhich  intersects  a 
given  line  at  any  angle. 

Let  {a\   a")  be  the  given  point,   and   {b'c',   b"c")   the  given   line  (Figs. 
211,  212). 

Bring  the  plane,  which  they  determine,  by  rabattement  around  a  hori- 


zontal  {a'o\  a"o")  (Fig.  211),  or  a  vertical  (^7 'c',  a"o"^  (Fig-  -i-\  passing 
through  the  point  {a\  a").  During  rotation  the  point  remains  fixed,  and 
the  line  {b'c',  b"e")  assumes  the  position  {b,"e,",  b,\\').  Through  {a\  a") 
lead   two   lines  which   make   with   the    perpendicular  {a"/,",  a'//),   let   fall 


86 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


from    this   point   upon    the    line  {b^"c,",  b,'c,'),   the  complement  of   the    re- 
quired angle,  and  counter-rotate  the  plane. 

The  points  (r/',  r,')  and  {d^",  d^)  are  now  projected  at  {e" ,  e')  and 
\d" ,  d')\  joining  them  with  («',  a"),  the  lines  {a"e",  a'e')  and  {a"d",  a'd') 
are  the  lines  souirht. 


II.      ANGLES    BETWEEN    LINES    AND    PLANES. 


213.  The  angle  between  a  line  and  a  plane  is  measured   by  tJie  angle     -^ 
which  the  line  makes  with  its  projection  on  that  plane.  "\ 

214.  Problem. —  To  determine  the  angle  between,  a  line  and  a  plane. 


Fig.  313 


p. 

.314 

1    \ 

/ 

(j:\a'   \ 

\ 


Let  {a'b\  a"b")  be  the  given  line,  and  {VM,  HM)  the  plane  (Fig.  213.) 

From  any  point  («',  a")  of  the  line  let  fall  a  perpendicular  upon  the 
plane,  and  find  the  angle  c"a^'d"  between  these  two  lines  (Art.  208). 
The  complement,  d"a^"e",  of  this  angle  is  the  required  angle. 

When  the  given  line  is  a  vertical  (Fig.  214),  pass  through  the  line  a 
plane  perpendicular  to  the  horizontal  trace  HM  of  the  given  plane ;  the 
two  planes  intersect  in  a  line  {a'b',  a"b"),  which  contains  the  projection 
of  the  given  line  upon  the  plane  M.  The  angle  between  the  line  and 
its  projection  may  be  determined  by  rotating  {a'b',  a"b")  around  {c'd', 
e"d")  as  an  axis  until  it  is  brought  parallel  to  V,  when  a  is  the  required 
angle. 

When  the  given  line  is  the  ground-line  (Fig.  215),  let  fall  from  any 
point  X  a  perpendicular  to  the  given  plane  {HM,  VM),  and  find  its 
piercing-point  {c,  c")  thereon ;  o'c',  o"c"  are  the  projections  of  the 
ground-line  upon  the  plane  M.  But  o" x"  is  the  hypothenuse  of  a  triangle, 
right-angled  at  {c' ,  c"),  of  which  0" c"  is  the  horizontal  projection  of  the 
base,  and  x"c"  of  the  perpendicular ;  henee  it  may  be  inscribed  in  a  semi- 
circle described  around  x"o"  as  a  diameter.  Rotating  the  semicircle  into 
H,  the  point  c  falls  at  r,",  and  x"o"c^'  is  the  angle  sought. 

215.  Problem. —  To  pass  through  a  given  point  a  line  making  any  angle 
'T.vith  a  plane. 

Let  {a',  a")  be  the  given  point,  {VM,  HM)  the  plane,  and  a  the. 
angle  (Fig.  216). 

From  {a',  a")  let  fall  a   perpendicular  upon  J/,  find   its   piercing-point 


«7 


{b',  b")   thereon,  and    draw   through    it 
a  horizontal  \b'c\  h"c")  or   a   vertical. 


any  Hne  of   the    plane,   preferably 
The    two  lines  drawn   through   a. 


and    making    the    required    angle    with   {b'c\  b"c'  ), 
The  problem  is  then   resolved  into  a  special  case 


are  the 

required  lines 

of  Art. 

2  I  2. 

Fiy.Gio 

^ 

^ 

'  w- 

2l6. 

urc    the 
lines  of 


-Those   lines   in   a   plane   whicii 


fj'f/cs  of  Greatest    Declivity 

greatest  possible  angle  with  any  second  plane  are  termed  the 
greatest  declivity.  As  they  are  perpendicular  to  the  line  of  in- 
tersection between  the  two  planes  (Art.  71),  it  follows  that  those  which 
measure  the  greatest  angle  to  H  are  perpendicular  to  the  horizontals, 
and  those  which  measure  the  greatest  angle  to  V  are  perpendicular  to 
the  verticals.  Hence  any  i)lane  cutting  a  given  plane  perpendicular  to  its 
horizontals  or  verticals  will  determine  respectively  the  lines  of  greatest 
declivity   to   //  or   \\ 

217.    Pkor.LEM. —  To  determine  a  plane  when  its  line-  of  greatest  declivity 
is  given.  ^.  ""^  '-''     7^ 

Let  {a'b',  a"b")  be  the  given   line  measuring  tH^   greatest    declivitv  to 
//(Fig.  217).  '  ^  '        \, 

perpendicular  to  the   horizontals,  lead   through   any 


point,  as  {b' ,  b"),  of  the  given  line  a  horizontal  of  the  plane ;  its  hori- 
zontal projection  e"b"  is  perpendicular  to  a"b".  By  means  of  these  two 
intersecting  lines  the   plane  can  readily   be  determined  (Art.  85 1. 

In   a   like    manner  when    the    measure   is  to   V  (Fig.  218),   the  vertical 
projection  of  the  given  line   is  perpendicular  to  the  verticals. 


88  ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 

2l8.  Problem. —  To  determine  the  projections  of  a  line  %vJicn  its  incliiuz^ 
tions  to  the  coordinate  planes  are  given  (Fig.  219). 

Let  a  and  §  be  the  given  angles,  marking  respectively  the  inclina- 
tions to  H  and    V. 

Take  any  point  {a' ,  a")  and  assume  that  the  required  line  passes 
through  it,  thus  forming,  with  its  horizontal  projection,  a  triangle  which 
is  right-angled  at  a".  By  rabattcment  around  a'a"  it  is  shown  in  its  true 
size  at  {a'a"b^').  But  db^'  is  likewise  the  hypothenuse  of  the  right-angled 
triangle  whicK  the  line  in  space  forms  with  its  vertical  projection ;  hence. 


constructing  this  triangle,  a'h'  is  the  length  of  the  •  vertical  projection. 
With  d  as  a  centre,  and  with  a  radius  equal  to  a'b,',  describe  an  arc 
cutting  GL  at  b' ;  db'  is  the  vertical  projection  of  the  line  sought.  With 
a"b"  as  a  radius,  and  with  a"  as  a  centre,  describe  a  second  arc  intersect- 
ing the  ordinate  let  fall  from  b'  in  b"  \  d'b"  is  the  horizontal  projection 
of  the  line. 

III.      ANGLES    BETWEEN    PLANES. 

219.  The  angle  between  two  planes  may  be  reduced  to  the  case  of 
the  angle  between  two  lines ;  for,  if  from  any  point  assumed  within  the 
angle  perpendiculars  be  drawn  to  the  two  planes,  the  angle  measured 
by  these  perpendiculars  will  be  the  supplement  of  the  angle  of  the  planes. 

The  angle  is  likewise  directly  measured  by  the  two  lines  which  are 
cut  by  an  auxiliary  plane  led  perpendicular  to  the  line  of  intersection 
between  the  two  planes. 

220.  Problem. —  To  determine  the  angle  betzueeti  two  planes. 

Let  {HM,  VM)  and  {HL,  VL)  be  the  traces  of  the  given  plane  (Fig. 
220). 

Find  the  projection  {db',  d'b")  of  the  line  of  intersection,  and  pass  a 
plane  perpendicular  to  this  line,  giving  the  horizontal  trace  d"c"e"  per- 
pendicular to  d'b" .  This  auxiliary  plane  cuts  the  given  planes  in  lines 
which  form  with  d"c"e"  a  triangle  whose  vertex  lies  in  the  line  {a'b' ,  d'b")\ 
but,  as  this  vertex  is  common  to  the  auxiliary  plane  and  to  the  plane 
projecting  ab  in  d'b" ,  it  is  a  point  of  their  line  of  intersection  which 
passes   through  c"  and   is   perpendicular   to   {a'b',  d'b").      Hence,   by   the 


ANGLES. 


89 


rabattcment  of  ab  around  a"b'\  the  distance  c"f^\  found  by  drawing  a 
I^erpcndicular  from  c"  to  a"b^',  determines  the  revolved  projection  of  the 
vertex  //'  of  the  triangle.  /  Laying  off  this  measurement  from  c"  upon 
c"a",  and  drawing  f^'d"  and  f.,"c",.  the  included  angle  a  is  the  angle 
sought. 

In   Fig.  221   the   horizontal  traces  coincide;    hence  the    common    trace 
is   the    line   of   intersection    between   th^   two   planes,   and    the   auxiliary 


cutting  plane  passed  perpendicular  to  it  is  likewise  perpendicular  to  //. 
By  rabattcment  around  d"b"  the  lines  of  intersection  d"c^"  and  d"b^"  are 
determined,  and   with   them  the  included  angle  a. 

In  Fig.  222'  the  horizontal  traces  are  parallel;  hence  their  line  of 
intersecti(Hi  {a'b\  a"b")  is  parallel  to  H,  and  the  auxiliarv  cutting  plane 
is  perpendicular  to  it.  By  rabattcment  around  d"e"  the  triangle  and  the 
required  angle  01  are  determined. 

In  Fig.  223  the  traces  of  the  respective  planes  form  continuous  lines, 
and  their  line  of  intersection  {a'b\  a"b")  lies  in  a  plane  perpendicular  to 


GL.  Pass  an  auxiliary  cutting  plane  through  the  ground-line  and  per- 
pendicular to  {a'b\  a"b"),  intersecting  the  given  planes  in  the  sides  of  a 
triangle,  the  base  of  which  is  xj',  and  the  line  ab  in  a  point  which  is 
the  vertex  thereof.  Rotate  ab  into  V  around  b'b"  as  an  axis,  and  from 
b"    let    fall    the    perjicndicular   b" 


90 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


volved  projection  of  the  vertex.  Lay  off  from  b"  the  distance  b"c^  upon 
b"a" ,  and  a  is  the  angle  sought. 

221.  Problem. —  To  determine  the  angles  which  a  given  plane  makes  zvith 
the  coordinate  planes. 

Let  {VM,  HM)  be  the  given  plane  (Fig.  224). 

The  traces  being  the  lines  of  intersection  between  the  given  plane 
and  the  coordinate  planes,  the  cutting  planes  {HL,  VL)  and  {VN,  HN) 
respectively  determine  the  inclinations  to  H  and  V.  By  the  rabattement 
of  the  triangles  around  the  traces  HL  and  VN  the  required  angles  or 
and  /?  are  determined. 

In  Fig.  225  the  rabattement  has  been  effected  around  VL  and  HN 
respectively. 


222.  Problem. —  To  determine  the  planes  whicli  contain  the  line  of  inter- 
section between  two  planes  and  zvhich  bisect  their  angles. 

Let  the  planes  be  given  as  in  Fig.  226. 

Find  the  angle  between  the  two  planes  (Art.  220),  and  bisect  the 
ano-les  by  means  of  the  projections  {h^'g")  and  {h^'T)  Restoring  the 
triangle  g"K'e"  to  its  original  position,  the  bisecting  lines  become  lines 
of  the  bisecting  planes.  During  this  counter-rotation  the  piercing-points 
{g"  1  S')  ^^"^d  ^^" ■>  /')  ^^'^^^"''  fixed  and  serve,  in  connection  with  the  pierc- 
ing-points (//',  b'')  and  {a",  a')  of  the  line  of  intersection,  to  determine  the 
bisecting  planes. 

223.  Problem. —  To  find  the  traces  of  a  plane  zuhen  its  inclination  to  the 
coordinate  planes  is  given. 

As  the  lines  of  greatest  declivity  measure  the  inclinations  to  the 
planes,  assume  that  the  auxiliary  planes  cutting  these  lines  pass  through 
a  common  point  a"  in  GL  (Fig.  227).  Again,  the  auxiliary  planes,  being 
perpendicular  to  the  required  plane,  intersect  in  a  line  a"c,  which  is  also 
perpendicular  to  that  plane  and  to  the  lines  of  greatest  declivity  {b'c 
and  e"c)  passing  through  its  foot  c. 

Assume,  then,  the  point  {a',  a")  as  a  point  of  the  required  plane 
(Fig.  228);  and  by  rabattement  around  a'a"  determine  the  triangle  a'a"e^', 
whose  hypothenuse  aU\'  measures  the  greatest  declivity  a  to  H.     Draw 


AXG/.ES.  91 

a"c^    perpendicular  to  ^V/,  and   from  a"  as  a  centre,  and  with  a  radius 

equal  to  a"i\',  describe  an  indefinite  arc;    it  is  evident  that  the  horizontal 
trace  of  the  plane  will  be  tana^ent  to  it. 


KiS.  22e 


The  line  a"c  ni  space  (Fig.  227)  makes  with  the  trace  ig'g")  an 
angle  equal  to  fi  \  hence,  if  the  angle  c^a'd^  (Fig-  228)  be  made  equal 
to  /^,  a"d^  will  equal  g'g"\  and  by  describing  an  arc  from  a"  with  a"d^ 


as  a  radius,  the  point  g"  will  fall  in  the  ordinate  drawn  from  g'.  The 
horizontal  trace  passes  thnnigh  g"  and  is  tangent  to  the  arc  f'V,'  at  c" , 
iuid  the  vertical  trace  passes  through  the  assumed  point  a'. 


92 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


CHAPTER    X. 

CHANGE   OF   POSITION    BY   COMBINED    MOTIONS. 

224.  In  applying  the  methods  of  rotation  to  other  than  simple  geo- 
metrical surfaces,  the  graphical  constructions  will  be  simplified  by  the 
employment  of  an  additional  motion  from  place  to  place,  whereby, 
instead  of  having  the  various  projections  which  result  from  the  change 
in  position  overlapping  and  difficult  to  disentangle,  they  are  separated, 
and  consequently  more  readily  seen  and  apprehended.  The  motion  by 
which  this  change  is  effected  is  termed  a  motion  of  translation  or  trans- 
position. 

225.  Problem. — Given  a  triangle,  required  to  find  its  projections  luhen  the 
axis  is  made  to  incline  to  both  planes  of  projection. 

(i)  Project  the  triangle  in  a  simple  position  ;  for  example,  with  its 
axis  perpendicular  to  H  (Fig.  229,   i). 

(2)  Let  the  triangle  so-  placed  be  moved  from  its  first  position  towards 
the  right,  each  point  describing  a  parallel  to  GL.  When  far  enough 
removed  to  avoid  interference  with  the  previous  position,  rotate  it 
around  c^c^'  until  the  axis  assumes  the  required  position  to  H.  It  is 
evident   that,  as   the   object   has   not   changed    its   position  to   F,  its  ver> 


tical  projection  in  the  second  position  must  be  an  exact  counterpart  of 
the  vertical  projection  in  the  first,  the  sole  difference  being  in  the  in- 
clination to  GL. 

Hence,  measure  from  r/  (Fig.  229,  2)  a  distance  r/^/  equal  to  c'a' , 
making  with  GL  the  complement  of  the  given  inclination  of  the  axis  to 
H,  and  complete  the  second  vertical  projection  a'b^c^.  rAs  each  point, 
during  the.  combined   motion   of   translation    and    rotation,   has   preserved 


CHANGE    OF  POSITION  BY  COMBINED  MOTIONS. 


93 


its  original  distance  from  V,  the  horizontal  projection  of  any  point  will 
be  found  at  the  intersection  of  the  ordinates  drawn  from  the  second 
vertical  and  first  horizontal  projections.  Thus,  a^"  falls  at  the  intersec- 
tion of  a"a^"  and  a^'a,",  b^"  at  the   intersection  of  b"b^"  and  b^'b^",  etc. 

(3)  Move  the  triangle  towards  the  right  and  rotate  it  around  a  ver. 
tical  line  passing  through  r,"  until  the  axis  has  assumed  the  required 
position  to  V  \  the  horizontal  projection  a^'b^'c^'  will  be  precisely  the 
same  as  that  in  the  second  position,  the  difference  being  in  the  inclination 
to  GL  (Fig.  229,  3).  During  the  combined  motions  of  translation  and  ro- 
tation each  p(jint  has  preserved  its  distance  from  H  unaltered  ;  hence  the 
vertical  pnjjections  of  the  points  will  be  found  at  the  intersections  of 
the  ordinates  drawn  from  the  horizontal  projection  of  the  third  position 
and  the  vertical  of  the  second.  Thus,  a^  falls  at  the  intersection  of 
a^'a„^  and  a^a^,  b^  at  the  intersection  of  b^'b^  and  b^b.^,  etc. 

226.    Problem. —  To  project  a  cube  in  a  doubly  oblique  position. 

(i)    Find  the  projection  in  a  simple  position  (Fig.  230,   i). 

(2)  By  a  motion  of  translation  bring  the  cube  so  placed  to  the  right 
of  the  first  position  and  rotate  it  around  ^/V,'  until  it  assumes  the  re- 
quired position  to  H.  During  these  combined  motions  the  object  has 
not  changed  its  position  to  V\  hence  the  vertical  projection  of  the 
second    position   (Fig.   230,  2)   will    be    precisel}'   the  same  as  that  of    the 


iirst.     Draw  r,'^?,'.  making  with   GL  the  complement  of  the  given  inclina- 
tion of  the  \axis  to  //.  and   complete   the   vertical  projection. 

Tlje  horizontal  projection  is  determined  as  in  Fig.  229,  2.  by  the  in- 
tersections  of  the  ordinates  </,'«,"  and  a"a^\  b^'b^"  and  b"b^\  etc.,  the 
outermost  lines  and  the  upper  base  being  represented  in  full  lines;  but 
c,"b,",  c,"d,"  of  the  lower  base  and  the  edge  f.'V,"  are  drawn  in  dotted 
lines,  as  they  are  hidden  from   view   in  looking  downwards. 


94 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


(3)  Move  the  cube  so  placed  towards  the  right  and  rotate  it  arouncf 
a  vertical  line  passing  through  r,"  until  its  axis  shall  be,  for  example, 
in  a  plane  perpendicular  to  GL  (Fig.  230,  3).  As  the  solid  by  these 
motions  changes  its  position  only  to  V,  each  point  moving  horizontally, 
in  a  circle  and  maintaining  its  height  above  H  unaltered,  therefore  the 
horizontal  projection  of  the  third  position  must  be  an  exact  countei  part 
of  that  of  the  second,  the  sole  difference  being   in  the   inclination  to   GL. 

The  vertical  projection  will  be  found,  as  in  Fig.  229,  3,  at  the  inter- 
sections of  the  ordinates  («/'«/.  ^iX')'  ^^-I'^'v  b^'l\'),'  Qtc,  drawn  respect- 
ively from  the  horizontal  projection  of  the  third  position  and  the  vertical 
of  the  second. 

227.  Problem. —  To  determine  the  projections  of  an  oblique  hexagonal 
Pyramid  in  a  doubly  oblique  position. 

(i)    Let  Fig.  231,   I   be  the  surface  in  its  simplest  position. 

(2)  Move  the  solid  to  the  right  and  incline  the  axis  until  it  assumes 
the  angle  a  to  H.  During  "this  rotation  around  a^'a^  each  point  of 
the  surface  describes  an  arc  parallel  to  V,  the  measure  of  which  is 
equal  to  /?,  the  difference  between  the  inclination  of  the  axis  to  H  in 
its  first  and   in  its  second   positions. 

Draw  rtjV/  (Fig.  231,  2)  equal  to  a'd'  and  making  with  GL  the  angle 

Eie.,S31 


§\  set  off  from  a^  the  distances  a^b^,  b^'c^',  etc.,  equal  to  a'b',  b'c',  etc., 
and  from  0,'  draw  the  axis  o^'z\',  making  the  required  angle  to  the  base. 
Complete  the  new  vertical  projection  by  connecting  the  vertex  t\'  with 
the  angular  points  of  the  base. 

The    horizontal  projection  is  found,   as  in   preceding    cases,   by   means 
of   the  intersecting   ordinates   drawn    from   the    second  vertical   and    first 


CHANGE    OF  POSITION  BY  COMBINED  MOTIONS. 


95 


horizontal  projections.  The  vertex  fulling  within  the  base,  the  entire 
jjrojection  now  becomes  visible. 

(3)  Transpose  the  surface  as  before,  noting  that  it  has  not  changed 
its  position  to  H,  and  that,  as  a  consequence,  its  horizontal  j)r(ijection  is 
precisely  the  same  as  that  of  the  second  position,  the  sole  difference 
beinjj;  in  the  inclination  of  the  axis  to  the  ground-line. 

The  vertical  projecticjn  is  determined,  as  before,  by  means  of  the 
ordinates  drawn  from  the  horizontal  projection  of  the  third  and  vertical 
projection  oi  the  second  position.  The  vertex  being  turned  (jutwards, 
the  entire  base  becomes  visible,  while  the  edges  T'./rz/  and  t'///  are  concealed 
and  marked  in  dotted  lines.  A  test  of  the  accuracy  of  the  construction 
will  be  found  in  tlic  parallelism  of  th(jse  lines  which  are  parallel  in 
space. 

228.  The  i)rojection  of  an  object  in  a  doubly  oblique  position  may 
be  determined  without  necessarily  resorting  to  the  first  two  auxiliary 
positions,  by  the  employment  of  the  methods  of  rabattcmcnt  and  sup- 
plementary planes.  The  constructions  may  likewise  be  modified  by  the 
consideration  of  the  fact  that  the  real  lengths  of  lines  in  space  stand 
in  a  fixed  proportion  to  their  projections.  A  scale  of  reduction  may 
thus  be  formed  and  applied  to  the  solution  of  the  third  position.  In 
this  manner  are  formed  those  axonometrical  constructions — of  which  Iso- 
metrical  Projection  is  an  illustration — the  development  of  which  is  a 
special  application  of  Descriptive  Geometr}-. 

229.  PkOHLEM. —  To  determine  tJie  projections  of  a  right  cylinder  in  a 
doubly  oblique  position. 

Fig.  233 

2 


By  a  rabattenient  of  the   lower  base  the  first  position  may  be  omitted 
(Fig.  232,  2). 


-96  ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 

The  axis  being  parallel  to  V,  the  vertical  projection  of  the  surface  is 
a  rectangle  the  base  of  which  is  equal  to  the  diameter  of  the  circle,  and 
the  altitude  that  of  the  given  cylinder.  By  rabattement  around  a  ver- 
tical of  the  plane  of  the  base — preferably  {a'b\  a"b"),  which  passes 
through  the  centre — the  circle,  projected  in  its  true  shape,  may  be 
divided  into  any  number  of  parts,  and  the  intermediate  points,  ^/,  d^',  e^, 
etc.,  together  with  their  distances  from  the  axis,  readily  determined. 
By  counter-rotation  the  horizontal  projections  c" ,  d" ,  e" ,  etc.,  are  found 
by  measuring  on   either  side   of  a"b"   distances   equal   to  c'c^,  d'd^'y  and 

The  constructions  for  the  third  position  do  not  differ  from  those 
already  indicated  in  the  preceding  cases. 

230.  Problem. —  To  determine  the  projections  of  a  right  cone  when  the 
inclination  of  its  axis  to  the  coordinate  planes  is  given. 

The  projections  may  be  found  without  recourse  to  either  of  the  first 
two  auxiliary  projections  (Fig.  233). 


p^g.  333 


\ 


,-'  \"       W  /  1    \u   I  /I    // 


Let  the  projections  of  the  axis  {a'c\  a"c")  be  determined  as  m  Art, 
218.  As  the  base  of  the  cone  is  at  right  angles  to  the  axis,  the  plane 
passing  through  the  lower  extremity  of  the  axis  and  perpendicular  to 
it  coincides  with  the  base.  Hence  the  problem  is  resolved  into  the 
case  of  a  surface  which  stands  upon  a  given  plane.  Art.  197. 

Pass  through    the    extremity  {c' ,  c")  a  plane  M  perpendicular   to   the 


CHANGE   OF  POS/-JIOX  BY  COMBINED  MOTIONS.  97 

line  {a'c',  a"c") ;  the  horizontal  {c"p",  c'p')  giving  the  piercing-point  (/'. 
p")  will  be  sufficient  to  determine  the  traces  [VM,  HAI),  which  arc- 
respectively  perpendicular  to  the  projections  of  the  axis.  By  rabattcvunt 
of  the  plane  M  around  the  horizontal  the  circle  of  the  base  is  projected 
horizontally   in  its  true  size  e^"g"i,"b". 

Assume  a  new  vertical  plane  L  perpendicular  to  the  horizontal  {b'c', 
b"c");  its  horizontal  trace  ML  is  perpendicular  U)  b"c".  Find  the  new 
vertical  projection  (/,V/)  of  the  circle,  and  the  line  of  greatest  declivity 
0(^/.  or  the  intersection  between  the  two  planes  L  and  J/,  and  counter- 
rotate  the  plane  M.  During  this  rotation  the  plane  of  the  circle  re- 
mains perpendicular  to  the  new  vertical  plane,  and  the  projection  {i^i\) 
assumes  the  position  i^c^,  each  point  having  described  an  arc  parallel 
to  L  and  measured  by  the  angle  i^b^i^.  Letting  fall  ordinates  from 
the  different  points  in  this  position,  the  horizontal  projections  i" ,  b",  c'\ 
g'\  etc.,  are  found  in  the  lines  passing  through  the  points  /,",  b'\  i\",g", 
etc.,  and   perpendicular  to  the  horizontal  {b"c",  b'c'). 

The  vertical  projections  i',  b',  e' ,  g',  etc.,  may  be  found  either  by 
means  of  the  horizontals  led  through  the  individual  points  determined 
in  the  horizontal  projection  or  by  means  of  the  lines  of  greatest  de- 
clivity e'c',  e"c",  etc. 


9^  ELEMENTS  OF  DESCRIPTIVE   GEOMEl^RY. 


CHAPTER   XL 

SECTIONS. 
I.      ELEMENTARY    SECTIONS. 

231.  A  plane  may  assume  two  general  positions  to  a  surface:  it 
may  either  touch  or  cut  it,  or,  in  other  words,  it  may  be  either  a 
tangent  or  a  secant.  In  the  first  case,  as  a  general  rule,  the  surface 
lies  wholly  on  one  side  of  the  plane,  and  in  the  second  case  it  is 
divided  by  it. 

232.  The  line  in  which  the  cutting  plane  intersects  the  surface  is 
termed  the  line  of  intersection,  and  is  common  both  to  the  surface  cut 
and  to  the  cutting  plane ;  hence,  whatever  the  nature  of  the  surface, 
the  line   of  intersection  must   be  a  plane  line. 

233.  The  character  of  this  line  is  affected  by  two  considerations  :  {\) 
by  the  nature  of  the  surface  itself-,  (2)  by  the  position  of  the  cutting 
plane.  It  is  evident  that  a  section  of  the  sphere  must  be  wholly  curvi- 
linear and  diminish  in  size  as  the  plane  recedes  from  the  centre,  while 
the  section  of  a  pyramid  must  be  wholly  rectilinear  and  diminish  in  size 
as  the  plane  approaches  the  vertex. 

234.  As  influenced  by  the  nature  of  the  surface,  the  section  may, 
therefore,  be  curved,  rectilinear,  or  a  combination  of  these  two ;  as 
mfluenced  by  the  position  of  the  secant  plane,  the  section  may  vary 
as  the  plane  passes  through  a  centre,  an  element,  an  axis,  a  diameter, 
etc. 

235.  Longitiidinal  Sections. — In  every  prismatic  or  cylindric  surface,  if 
the  secant  plane  passes  through  an  edge,  rectilinear  K. 
element  or  axis,  it  will  cut  the  longitudinal  sec- 
tion composed  of  parallel  rectilinear  elements;  and 
if  the  surfaces  be  limited  by  bases,  as  in  the  geo- 
metrical solids,  the  section  will  be  a  parallelogram 
or  rectangle. 

236.  In  every  pyramidal  or  conic  surface,  if 
the  secant  plane  passes  through  the  vertex,  it  will 
cut  the  longitudinal  sections  composed  of  recti- 
linear elements  intersecting  in  the  vertex  ;  and  if 
the  surfaces  be  limited  by  bases,  the  section  will  be  a  triangle,  isosceles. 


SF.CnOA'S. 


99 


right  or  scalene  as  the  surface  is  right  or  oblique.  In  the  oblique 
cylinder  and  cone  any  longitudinal  plane  passed 
tiirough  the  surface,  at  right  angles  to  the  plane 
which  contains  the  axis  and  the  line  measuring 
the  altitude,  will  cut  the  rectangle  and  isosceles 
triangle  respectively. 

237.  Meridian  Scrtio)is. —  In  surfaces  of  rotation 
the  longitudinal  section  becomes  the  meridian  sec- 
tion, and  the  secant  plane  the  meridian  plane.  They 
are  determined  by  passing  through  the  axis  a  secant 
plane,  which  thus  coincides  with  and  cuts  a  gen- 
eratrix of  the  surface.  Hence  the  meridian  sections 
of  surfaces  of  rotation  are  equal. 

238.  It  has  already  been  shown  that,  in  generat- 
ing any  surface  of  rotation,  each  point  of  the 
moving  line  describes  a  circle  whose  plane  is  per- 
pendicular to  the  axis  and  whose  centre  is  in  it ; 
hence  any  section  perpendicular  to  the  axis  of  such 
a  surface  must  always  coincide  with  one  of  these 
circles,  termed  collectively  the  parallels.  Except  in  the  case  of  cvlin- 
dric  surfaces,  these  parallels  are  unequal,  the  largest  being 
known  as  the  circle  of  the  equator,  the  smallest  the 
circle  of  the  gorge. 

As  evei-y  meridian  section  is  cut  by  a  plane  which 
passes  through  the  axis  and,  hence,  through  the  centres 
of  the  parallels,  it  necessarily  divides  both  parallels  and 
surface    into  two   equal    parts. 

239.    Perimetrieal    Seetions. — When    the    secant    plane    in- 
tersects   the    consecutive    generatrices    of    a    surface,    the 
section   thus  obtained   is  termed   a  perimetrieal  section. 
In    the    cylindric    and    prismatic    surfaces   such    a    section   is  elosed ;   in 
the    pyramidal    and    conic    it    may    be   open.      Taking   an    oblique    circular 
cylinder  by    way  of  illustration,  it   will   be  found   that — 
(i)    The  parallels  are  equal  to  the  base. 

(2)  All  sections  cut  by  planes  which  intersect  the 
surface  at  an  angle  i^,  equal  to  the  inclination  of 
the  axis  to  the  base  but  in  the  opposite  direction, 
are  equal  circles,  the  sub-eontrary  seetions. 

(3)  Every  section  between  a  parallel  and  its  sub- 
contrary  will  be  an  ellipse  with  its  transverse  axis 
cHjual  to  the  diameter  of  the   base. 

(4)  Every  section  beyond  the  i)arallcl  and  its 
sub-contrarv  will  be  an  ellipse  with  its  eonjugate 
axis  equal   to   the   diameter  of  the   base. 

(5)  The   smallest   section    will    he   cut    j)erpendicular  to   the  axis. 


ELEMEXTS   OF  DESCRIPTIVE    GEOMETRY. 


be 


240.  In  the  pyramidal  and  conic  surfaces  the  section  may  be  open 
or  closed,  but  all  sections  cut  by  parallel  planes  are  similar. 

Taking   an    oblique    circular    cone    by    wa)^    of    illustration,    it 
found  that  the  secant   jUane   may  intersect : 

(i)  All  the  rectilinear  elements  of  one  nappe 
only,  giving  a   closed  section. 

(2)  All  the  rectilinear  elements  of  one  nappe 
except  that  element  to  Avhich  the  plane  may  be 
assumed   parallel,  giving  an  open  curve. 

(3)  All  the  rectiHnear  elements  of  both  nappes 
except  the  two  to  which  the  plane  may  be  assumed 
parallel,  giving  an  open  curve  of  tzvo  branches. 

(4)  At  an  angle  equal  to  the  inclination  of  the 
axis  to  the  base,  but  in  an  opposite  direction, 
giving   the   circular  sub-contrary  sections. 

In  the  first  case  the  section  will  always  be  elliptical,  in  the  second 
parabolic,  in  the  third  hyperbolic,  whilst  all  those  sections  cut  between 
the   second   and  third   will  also  be  hyperbolic. 

241.  In  the  greater  number  of  surfaces  the  sections  cut  by  the  peri- 
metrical  and  longitudinal  planes  coincide  with  the  generatrices  and  direc- 
trices, and  hence  are  termed  the  elementary  sections.  In  practice  they  serve 
as  most  useful    auxiliaries  in  the  determination  of   lines  of   intersection. 

In  prismatic  and  pyramidal  surfaces  the  edges  which  mark  by  their 
piercing-points  on  the  secant  plane  the  breaking-points  of  the  line  of  in- 
tersection are,  as  a  rule,  sufficient  to  determine  that  Hne. 

In  cylindric  and  conic  surfaces  the  piercing-points  of  rectilinear 
elements  upon  the  secant  plane  indicate  the  required  points  in  the  line 
of  intersection.  The  parallels,  where  their  employment  is  readily  avail- 
able, may  also  be  employed  to  the  same  end. 

In  surfaces  of  rotation  the  parallels  and  meridians,  in  warped  surfaces 
the  rectilinear  elements,  and  in  other  surfaces,  as  the  polyhedrons,  the 
edges    and    faces,    are    the    auxiliaries    employed    in  the  work. 

242.  As  the  section  is  common  to  the  secant  plane  and  to  the  sur- 
face cut,  any  line  of  the  section  must  be  likewise 
a  line  of  the  plane ;  hence  any  point  of  the  line 
of  intersection  may  be  regarded  as  a  point  of 
intersection  between  a  line  of  the  plane  and 
another  of  the  surface.  If,  then,  the  surface 
should  be  cut,  within  the  possible  hmits  of  its 
intersection,  by  auxiliary  planes,  the  secant  plane 
Avill  also  be  cut,  and  the  two  lines  thus  determined 
lying  in  the  same  plane  will  intersect  in  one  or 
two  points  of  the  line  of  intersection. 

243.  An  important  consideration,  therefore,  in 
order  to  facilitate  the  graphical  constructions  is 
the  selection  of  the  position  which  the  auxiliary 
cutting   planes   should    be    made  to  assume  so   as  to  afford    the    simplest 


ELEMENTj^liY.    S/^C7/0.V,S.-\  •  lOl 

sections    of    the    surface    and    the    most    available    arrangement    for    pro- 
jection. 

On  the  surface  itself  the  secant  planes  should  be: 

(^i)    For  prismatic  and  cylindric  surfaces  parallel  to  the  axes. 

(2)  For  pyramidal  and  conic  surfaces  passed  through  the  vertices. 

(3)  For  surfaces  of  rotation  made  to  cut  the  parallels. 

244.  Prohlem. —  Tlirougk  a  point  on  the  surface  of  a  cone  to  cut  a  loni^i 
tudinal  and  a  parallel  section. 

The  longitudinal  plane  passes  through  the  vertex  (t-',  v")  and,  hence, 
through    the    line    joining   the   given    point  {a\  a")  with    the    vertex  (Fig. 

-341- 

From    these  data  an  indefinite    number  of   sections   may  be  cut,  all  of 


which  are  determined  bv  planes  which  pass  through  the  foot  (^',  ^") 
of  the  line,  intersect  the  base  in  any  line  (^'V",  /f'c'),  and  cut  (c'v,  c"v")y 
thus  completing  the  triangle.  Onlv  one  parallel  can  be  cut.  which, 
being  parallel  to  //,  is  projected  thereon  in  a  circle  which  contains 
the    given    point. 

The  point  being  common  to  both  cutting  i)ianes  lies  at  the  intersec- 
tion (a'\  a')  of  their  respective  sections. 

245.  Prohlem. —  Through  a  point  on  the  surface  of  an  oblique  cylinder  to 
cut  a  longitudinal  and  a  parallel  section. 

Through  the  given  j)oint,  of  which  the  vertical  projection  a'  (Fig. 
235)  only  is  given,  lead  an  element  (//<',  b"c")  of  the  surface  and  deter- 
mine the  horizontal  projection  a".  As  an  indefinite  number  of  secant 
l)lanes  may  be  passed  through  this  element,  select  at  will  any  second 
element  ie'd',  e"d"),  and  complete  the  section  by  connecting  the  ex- 
tremities of  the  elements. 

But  one  jiarallcl  can  be  cut.  whose  centre  <g'g")  lies  in  the  axis  "nd 
whose   jirojections  contain   those  of   the  given  point. 


102  ei:emi':?7ts  pF' descriptive  geoa/etry. 

246.  Problem. —  To  divide  a  horizontal  prism  into  tzvo  equal  parts  by  a 
longitudinal  plane  ivJiich  makes  a  gi%>en  angle  zvith  the  horizontal  plane;  and 
to  remove  the  front  section  of  the  surface. 

Let  the  prism  be  octagonal  (Fig.  236). 

Assume  a  new  ground-line  G^L^  perpendicular  to  the  primitive 
ground-line,  and  determine  the  new  vertical  projection  of  the  surface. 
As  the  secant  plane  is  perpendicular  to  the  new  vertical  plane,  the  pro- 
jection of  the  section  falls  in  any  line  w/«/.  From  the  new  vertical 
projections  the  primitive  projections  of  both  surface  and  section  are 
readily  found  according  to  Art.   1 54. 

The  broken  lines  represent  the  portion  of  the  surface  removed. 
When  the  section  is  exposed  to  view  b}^  the  removal  of  any  part  of 
the  surface,  the  fact  is  indicated  by  shading  the  section. 


247.    Problem. —  To  cut  the  meridians  on  a  surface  of  rotation. 

As  the  surface  may  always  be  brought  by  methods  already  demon- 
strated into  a  position  in  which  the  axis  shall  be  perpendicular  to  H, 
assume  that  position  as  in  Fig.  237. 

The  secant  planes  are  then  perpendicular  to  H,  and  the  sections 
projected  on  it  in  right  lines,  as  a"b",  c"d",  etc.  The  auxiliarv  planes 
which  cut  the  parallels  on  the  surface  (Art.  242)  and  the  lines  on  the 
secant  planes  are  thus  horizontal,  the  points  of  intersection  a",  e",  f", 
€tc.,  between  these  two  sets  of  lines  marking  the  required  points  in  the 
line  of  intersection. 

The  vertical  projections  are  obtained  by  drawing  ordinates  from 
these  points  to  the  auxiliary  circles  in  which  they  are  contained.  The 
meridians  pass  through  them. 


SECTIONS. 


103 


II.      OBLIOUE    SFXTIONS. 


248.  Where  a  surface  is  cut  by  a  plane  which  docs  not  coincide  with 
the  primitive  elements,  the  section  is  oblique.  Analytical  Geometry 
determines  the  character  of  such  sections,  and  the  results  (jf  mathe- 
matical investigations  have  established  the  following  facts  with  regard 
to  them  : 

(1)  The  cylinder  gives  c/Iipsi's. 

(2)  The  cone,  ellipses,  parabolas,  hyperbolas. 

(3)  In  surfaces  of  rotation — 

(V?)    The  ellipsoid  gives  ellipses. 

(b)   The  paraboloid,  ellipses  and  parabolas. 

(e)    The  hyperboloid,  ellipses,  parabolas  and  hyperbolas. 

(4)  In  surfaces  of  the  second  order  any  plane  section  is  limited  by  a 
line  of  the  second  order. 

(5)  In  surfaces  of  the  second  order  parallel  sec- 
tions are   similar. 

249.  The  secant  plane  always  cuts  a  surface 
of  rotation  in  a  symmetrical  figure,  the  transverse 
axis  of  which  is  the  line  of  intersection  between 
the  secant  plane  and  the  meridian  plane  at  right 
angles  to  it.  The  point  or  points  in  which  this 
axis  pierces  the  surface  will  be  the  vertex  or  ver- 
tices of  the  section  cut. 

250.  In  determining  the  perimeter  of  the  oblique  section  the  ele- 
ments, edges  or  other  auxiliary  lines  of  the  surface 
may  be  employed ;  the  practical  requirements  of  the 
case  demand,  however,  that  the  choice  shall  lie  with 
those  lines  which  render  the  constructions  most  simple 
in  character. 

251.  A  system  of  auxiliary  planes  which  cut  a 
series  of  auxiliary  lines  on  the  surface  will  also  inter- 
sect the  plane  of  the  oblique  section  in  right  lines, 
which  are  secants  to  that  section.  These  secants 
])ierce  the  surface  in  the  points  in  which  the  oblique 
])laiie  crosses  the  auxiliary  lines  traced  upon  the  sur- 
face. 

252.  Problem. —  To  iuterseet  a  given  prism  by  a  plane  perpendieular  to 
the  vertical  plane. 

Let  the   prism   be  given  as  in   Fig.  238. 

By  the  conditions  of  the  problem  the  vertical  trace  of  the  secant 
plane  may  be  any  line  VM,  while  its  horizontal  trace  HM  must  be  per- 
pendicular to  GL.  The  section  lying  in  a  plane  perpendicular  to  T^ 
is  projected  on  that  plane  in  a  line  a'e'  in  the  vertical  trace:  and  as 
the  entire   prism   is  projected    horizontally  in   the  perimeter  of  the  base. 


104 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


it    is   evident   that   any    part   thereof,    as    the   section,    must    Hkewise    be 
projected  therein. 

The  true  size  of  the  section  may  be  determined  by  rotating  the 
plane  M  around  either  trace  or  around  one  of  its  verticals,  preferably 
that  which  intersects  the  axis.  The  latter  construction  is  usefully  em- 
ployed where  the  space  is  restricted. 


Fig.SSS 

a:— 

2), 

VM 

y^^^ 

/ 

\ 

'C' 

/ 
/ 

/ 

\ 

?>^ 

/                       / 

\  1 

f                     r 

^ 

\ 

a'. 

-< 

^ 

^^ 

y' 

X  \ 

^ 

' 

^ 4— 

— 

-^ 

•< 

253.    Problem. —  To  intersect  an  oblique  pyramid  by  a  plane  perpendicular 
to  the  horizontal  plane. 

Fig.  339 


Let  the  pyramid  be  given  as  in  Fig.  239. 

The  secant  plane  {VM,  HM)  being  perpendicular  to  H,  the  section 
it  cuts  is  projected  in  the  horizontal  trace,  HM,  in  the  line  b"g" ;  the 
points  in  which  the  trace  cuts  the  horizontal  projections  of  the  edges 
are  projections  of  the  breaking  points  of  the  line  of  intersection.  Hence 
the  vertical  projections  are  found  by  drawing  the  ordinates  from  these 
points  to  the  vertical  projections  of  the  edges  in  which  they  lie. 

The  true  size  of  the  section  may  be  determined  by  rabattenicnt,  as  in 
the  preceding  case,  or  around  a  vertical  of  the  plane  passing  through 
the  point  {g' ,  g"). 


SECTIONS. 


105 


254.    Problem. —  To  intersect  a  right  cylinder  by  a  given  plane. 

Let  the  cylinder  be  given  as  in  Fig.  240,  and  let  VM,  HM  be  the 
traces  of  the  given  plane. 

Cut  the  plane  and  the  surface  by  a  series  of  auxiliary  planes  (Art. 
251),  the  surface  in  rectilinear  elements  and  the  plane  in  verticals.  In 
accordance  with  these  conditions  the  auxiliary  planes  must  be  parallel 
to  V,  giving  the  traces  c"k",  b" I" ,  a"/i",  etc.,  and  cutting  the  cylinder  in 
elements  the  horizontal  projections  of  which  fall  in  c",  a",  b" ,  etc. 
Determine  the  vertical  projections  of  these  elements  e'e',  (I'd',' etc.,  and 
of  the  verticals  c'i-',  b'l',  a! h' ,  etc.,  and  their  points  of  intersection  mark 
points  in  the  line  of  intersection  sought.  Thus,  c'c'  the  element  inter- 
sects k'c'  the  vertical  at  tJie  point  c ,  b'b'  intersects  b'l'  at  //,  and  simi- 
larly  for  intermediate  points  of  the  section  e'f'a'b'c'd'. 

The  plane  of  the  upper  base  being  horizontal  cuts  the  secant  plane 
j\I  in  a  horizontal  {e"f",  e'f),  which  completes  the  limits  of  the  section. 

To  determine  the  true  size  of  the  section,  bring  it  by  rabattement 
around  one  of  the  horizontals,  as  e"f"  of  the  secant  plane. 


.SAO 


UL'\ 


255.    l^ROBi.EM. —  To  intersect  a  right  cone  by  a  given  plane- 
Let   the  cone  be  given  as    in   Fig.  241,  and    lot    the    plane  J/  be    per- 
pendicular to   V. 

Pass  through  the  surfaces  a  series  of  longitudinal  cutting  planes  at 
right  angles  to  \'.  Each  plane  cuts  on  the  cone  two  rectilinear  elements, 
as  {b"v" ,  b'v')  and  {h"v",  h'v'),  which  coincide  in  the  vertical  jin\iections, 
and  a  line  {h"b",  lib')  on  the  secant  plane  which  intersects  the  elements 
in  the  points  b  and  h  respectively.  A  succession  of  points  may  thus 
be    obtained    through    which    the    required   section  abcdefgh  passes.     For 


^°^  ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 

the    point  (.',  .")    however,  as    the  auxiliary  sections  coincide    in    projec- 
tion and,   hence  determine  no  point  of  intersection,  a  parallel   instead  of 


a  longitudinal  section  has  been  employed 

To  determine    the    section    in    its   true    size,  revolve   the   secant   plane 
around   its  vertical  trace    ViM.  ^ 

Fig.  243 


Fig.  242 


Fig.  242  exhibits  the  case  in  which  the  secant  plane  has  been  assumed 
to   be   parallel    to   an   element   of    the   cone.      The   section    which    is   the 


SECTIONS. 


107 


parabola  may  be  found  by  constructujiis    similar   to    those    of   the  preced- 
ing example. 

Should  the  secant  plane  be  taken  parallel  to  the  axis  (Fig.  243),  or  in 
any  position  intermediate  between  this  and  that  immediately  preceding; 
the  hyperbola  will  be  the  section  cut,  since  both  nappes  of  the  surfaces 
will  be  intersected.  In  these  cases  the  parallels  will  prove  the  most 
available  auxiliaries. 

256.    l^KOHl.EM. —  To  intersect  a  surface  of  revolution  by  a  given  plane. 

Let  the  surface  be  given  as  in  Fig.  244,  and  let  (FJ/,  HM)  be  the 
given  plane. 

By  Art.  249  the  axis  of  the  section  will  be  determined  bv  findino- 
the  intersection  {p'o ,  p"o")  between  a  meridian  plane  L  and  the  given 
plane,  the  two  being  at  right  angles  to  each  other.  The  points  in  which 
this  line  pierces  the  surface  are  the  vertices  of  the  curve  of  intersection 
as  well  as  the  highest  and  lowest  points  thereof. 


Ki''.244r 


To  determine  these  points  revolve  the  meridian  plane  around  the 
axis  of  the  surface  until  it  becomes  parallel  to  V\  the  line  ^p'o\  p"o") 
then  assumes  the  position  {p,"o".  p/o'),  in  which  a,'  and  r/  are  the  ver- 
tices revolved.  By  counter-rotation  (a\  a")  and  {c,  e")  indicate  the  ver- 
tices in  the  required   position. 

Now,  by  a  series  of  horizontal  auxiliarv  {)lanes  between  these  two 
limiting  points,  cut  parallels  on  the  surtace  and   horizontals  on  the  secant 


io8 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


plane  M\  the  intersections  of  these  two  fix  a  succession  of  points^ 
a,  b,  c,  .  .  .  through  which  the.  required  section  passes.  Thus  the  hori- 
zontal {s"b",  s'b')  cuts  the  ciixle  (//'//',  ////')  in  the  points  {b" ,  b')  {h" ,  //'), 
the  horizontal  {l"g",  !'g  ')  cuts  the  circle  {c-"g" ,  c'g')  in  the  points  (r",  c') 
{g" ,  g'),  etc.  To  determine  the  section  in  its  true  size,  bring-  it  by 
rabattement  around  a  horizontal,  as  {/'d' ,  f'd"). 

257.    Problem. —  To   intersect  a  pyramid  by  a  given  plane. 

Let  the  pyramid  be  given  as  in  Fig.  245,  and  let  {VM,  HM)  be  the 
given  plane. 

Pass  through  the  edges  a  series  of  auxiliary  cutting  planes  perpen- 
dicular to  V,  intersecting  the  surface  in  the  edges  and  the  secant  plane, 
M,  in  a  series  of  lines;  the  points  in  which  these  lines  intersect  the 
edges  are  the  piercing-points  of  the  latter  on  the  secant  plane  or  the 
breaking-points  of  the  line  of  intersection  sought. 


As   the    auxiliary  planes 


pass  through  the  axis,  their   intersections 


with  the  secant  plane  M  must  pass  through  the  piercing-point  (/',  p")  of 
the  axis  on  that  plane.  Thus  the  auxiliary  plane  passing  through  the 
edges  {v'f,  v"f")  and  {v'c',  v''e")  cuts  the  secant  plane  in  a  line  whose 
horizontal  projection  i"k"  intersects  the  edges  in  the  points  (/',  /")  and 
{c,  c").  Similar  constructions  determine  the  points  {a',  a")  and  {b',  b")^ 
whilst  the  points  {d',  d")  and  (/,  e")  may  be  found  by  passing  auxiliary 
planes  through  the  sides  of  the  base  in  which  they  lie. 

The  true  size  of  the  section  may  be  determined  by  rabattement  around 
the  line  {d'e',  d"e"). 


SECTIONS. 


109 


258.  Problem. —  To  intersect  an  oblique  cylinder  by  a  plane  at  right  angles 
to  the  axis. 

Let  the  cylinder  be  given  as  in  Fig.  246,  the  surface  extending  in- 
definitely  upwards.  By  the  conditi(jns  of  the  case  the  traces  (//J/,  VM) 
of  the  plane  must  be  perpendicular  to  the  projections  of  the  axis  of  the 
same  name.     The  section  thus  cut  is  termed  a  right  section  oi  the  surface. 

Pass  a  series  of  vertical  longitudinal  planes  cutting  elements  upon  the 
surface  and  parallel  right  lines  upon  the  secant  plane  M.  Thus,  the 
plane  passing  through  the  elements  whose  feet  are  a"  and  e"  cuts  the 
plane  M  in  a  line  whose  horizontal  projection  coincides  with  that  of 
the  elements.  One  point  of  its  vertical  projection  may  be  immediately 
found  at  o  ,  and  the  second,  ;/,  may  be  determined   by  means  of  a  hori- 


Ki'-.2-*0 


zontal  led  through  anv  point  /'  of  the  vertical  trace  of  the  plane. 
Hence  the  line  to  which  the  vertical  projections  of  all  lines  cut  on  the 
plane  M  by  the  auxiliary  planes  must  be  parallel  passes  through  0  and 
//'.  The  intersections  of  {p' n  ,  o"n")  with  the  elements  passing  through 
{a',  a")  and  {e\  e")  determine  the  points  (/'.  /")  and  {g',  g")  of  the  right 
section  sought.  By  a  similar  procedure  a  series  of  rectilinear  elements  and 
of  lines  parallel  to  (o'n,  o"n")  may  be  obtained,  whose  intersections  mark 
points  in  the  vertical  projection  of  the  curve  of  the  section,  from  which 
the  horizontal  projections  may  readily  be  determined  by  means  of  ordinates. 


no  ELEMENTS   OF   DESCRIPTIVE    GEOMETRY. 

To  determine  the  section  in  its  true  size,  bring  the  plane  M  by  ra- 
battemettt  around  its  horizontal  trace  HM. 

259.  Problem. —  To  pass  a  plane  through  a  given  line,  and  to  cut  a  great 
circle  on  a  sphere. 

Let  the  sphere  be  given  as  in   Fig.  247,  and  let  L  be  the  given  line. 

As  every  plane  cutting  a  great  circle  must  pass  through  the  middle  of 
tlie  sphere,  from  any  point  {p' ,  0")  of  the  given  line  lead  a  line  through  the 
middle  {c\  c")  and"~^ determine  the  traces  (VM,  HM)  of  the  plane  passing 
through  these  two  lines  (Art.  85).  The  meridian  plane  {VN,  HN)  drawn 
perpendicular  to  the  plane  thus  found  cuts  the  transverse  axis  of  the  section 
whose  extremities  mark  the  highest  and  lowest  points  thereof  (Art.  249). 


Thus  the  meridian  -plane  N  cuts  the  line  {p"c",  p'c')  on  the  plane  M, 
and  on  the  sphere  a  great  circle  whose  horizontal  projection  is  r"s". 
Rotate  the  plane  N  around  the  vertical  axis  of  the  sphere  until  it  is 
brought  parallel  to  V,  when  the  Hue  {p"c",  p'c')  assumes  the  position 
{Pi"^'\  A'O  ^nd  determines  the  points  h,'  and  //,  the  resolved  posi- 
tions of  the  vertical  projections  of  the  highest  and  lowest  points  re- 
spectively of  the  section  sought. 

By  counter-rotation  h,'  falls  at  //',  and  //  at  /'.  Now  by  a  series  of 
auxiliary  horizontal  planes  between  these  two  points  cut  parallels  on  the 
sphere  and  horizontals  on  the  secant  plane  M\  their  intersections  mark 
points  of  the  great  circle  required. 


JNTEKSECTWNS. 


CHAPTER    XII. 

INTERSECTIONS. 


I.      ELEMENTARY     INTERSECTIONS. 

260.  When  the  distance  between  two  surfaces  is  reduced  to  a  mini- 
mum they  merely  touch  or  are  tangent  to  each  other,  but  if  this  distance 
be  measured  by  a  negative  quantity  they  intersect.  In  such  a  position 
they  have  a  line  in  common,  termed  the  line  of  intersection. 

261.  This  line  of  intersection  being  common  to  the  two  surfaces,  it 
follows  that  each  of  its  points  is  likewise  common;  hence,  in  determin- 
ing points  which  are  common  to  two  surfaces  the  line  of  intersection 
will  i)e  determined. 

262.  The  character  of  the  line  of  intersection  is  afTcctcd  bv  three  con- 
siderations :  the  nature  of  the  surfaces  themselves,  their  relative  posi- 
tions and  their  extension  beyond  the  initial  line  of  intersection.  Thus  it 
is  evident  that  the  rectilinear  surfaces  can  only  intersect  m  right  lines, 
double-curved  surfaces  in  curved  lines,  whilst  a  polyhedron  and  a  curved 
surface  may  present  a  line  which  is  partly  right  and  partly  curved. 

263.  It  is  to  be  noted  that,  other  things  being  equal,  a  change  in 
position  materially  modifies  the  character  of  the  line  of  intersection. 
Thus  a  pyramid  and  a  cone  whose  vertices  coincide  intersect  in  a  tri- 
angle, but  for  all  other  positions  the  plane  faces  of  the  pyramid  cut,  as 
a  rule,  perimetrical  sections  on  the  surface  of  the  cone. 

264.  The  variations  in  the  line  of  intersection  which  arise  from  the 
prolongation  of  the  surfaces  beyond  their  initial  intersection  may  be 
illustrated    by  means  of   the    cylinders   of  rotation. 

Assuming  the  axes  of  two  equal  cylinders  to  mtersect,  it  is  found  that  : 
(i)  Should  the  surfaces  stop  at  their 
first  contact,  the  line  of  intersection  will 
be  an  ellipse  or  oblique  section  of  either 
surface,  and  that  the  smaller  the  angle 
'«  between  the  axes  the  greater  the  trans- 
verse axis  of  the  ellipse.  In  such  a  case  the 
surfaces  are  said  to  intersect  by  meeting. 
(2)  Should  either  surface  be  extended 
bcytMul  the  first  line  of  contact,  the  intersection  assumes  the  form  of 
a   wedge,    whose    two  faces   are    semi-ellipses  which  respectivelv   increase 


112 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


and    diminish   with    the   angle    between    the    axes.      In    such    a    case    the 
surfaces    are    said    to    intersect    by  penetrating. 

(3)  Should  both  surfaces  be  extended  beyond  the 
first  line  of  contact,  the  line  of  intersection  is  com- 
posed of  two  ellipses.  In  such  a  case  the  surfaces 
are  said  to  intersect  by  crossing. 

265.  Requirements  for  Elementary  Intersections. — Two 
prisms  or  two  cylinders,  or  a  prism  and  a  cylinder, 
will  intersect  in  rectilinear  elements,  provided  their 
axes  or  edges  are  parallel. 

Two    cones    or    pyramids,    or    a    pyramid    and    a 
cone,  will  intersect  in  rectilinear  elements,  provided  their  vertices  coincide. 
Two  pyramids,  or  a  pyramid  and  a  prism,  whose  parallel  sections  are 
similar   and    whose    homologous   sides   are    parallel   will,  should    the  axes 
coincide,  intersect  in  one  of  the  parallels. 

Surfaces  of  revolution  whose  axes  coincide  intersect  in  parallels  or 
circles. 

266.  ProblExM. —  To  dispose  a  right  prism  and  right  cylinder  so  as  to 
intersect  in  rectilinear  elements. 

Let  the  axes  be  made  parallel  (Fig.  248) ;  then  the  faces  of  the  prism, 
being  parallel  to  the  axis  of  the  cylinder,  cut  that  surface  in  the  ele- 
ments {a",  a'a')  and  {b" ,  b'b'). 


Kims^ts 


irig.a4:9 


267.  Problem. —  To  dispose  two  cones  so  as  to  intersect  in  rectilinear 
elements. 

Let  the  vertices  coincide  (Fig.  249) ;  then  the  elements  {i>'a' ,  v"a") 
and  iv'b' ,  v"b"),  passing  through  the  vertices  and  the  points  of  intersec- 
tion of  the  bases,  are  common  to  the  two  surfaces,  and  hence  are  their 
line  of  intersection. 

268.  Problem. —  To  dispose  a  right  pyramid  and  right  prism  whose  re- 
spective parallels  are  similar,  so  as  to  intersect  in  a  parallel. 

Let  the  axes  coincide  (Fig.  250)  and  the  sides  of  the  bases  be  paral- 
lel ;    then  each   face   of   the    prism    cuts  a  face  of    the    pyramid   in  a  line 


IN  TER  SECTIONS. 


113 


which  is  parallel  to  a  side  of  its  base,  while  the  edges,  lying"  four  and 
four  in  the  same  vertical  planes,  intersect  in  the  breaking-points  of  the 
parallel  of  intersection. 


Fig.  aco 


269.  PR(^bli:.m. —  To  dispose  surfaces  of  revolution  so  as  to  intersect  in 
parallels. 

Let  the  axes  coincide  (Fig.  251);  then  any  meridian  plane  J/,  as  for 
example  that  parallel  to  V,  determines  points  on  the  apparent  contour 
through    which   the   required   parallels  pass. 

270.  Problem. —  To  deter  mine  the  circle  of  intersection  betzoeen  tico  spheres. 
Let  the  spheres  be  given  as  in  Fig.  252. 


Pass    a    vertical   meridian   jWanc,  J/,    through    the    centres   of  the   two 


114  ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 

surfaces,  and  determine,  by  the  rabattcmcnt  of  the  plane  around  its  hori- 
zontal trace  HM,  the  highest  and  lowest  points,  {a" ,  a')  and.(^",  b')  respect- 
ively, of  the  line  of  intersection.  Cut  any  number  of  auxiliary  parallels 
between  these  points  by  means  of  horizontal  planes ;  they  intersect,  two 
and  two,  in   points  of  the  line  of  intersection  sought. 

II.      PLANE    LINES    OF    INTERSECTION. 

271.  If  the  generatrices  of  two  surfaces  be  made  to  follow  a  common 
directrix,  the  surfaces  thus  generated  will  intersect  each  other  in  this 
line. 

272.  The  character  of  the  plane  line  of  intersection  will  be  affected 
by  the  following  general  limitations :  first,  that  the  surfaces  admit  of 
identical  sections ;  second,  that  they  are  collinearly  disposed — that  is,  have 
their  homologous  elements  in  the  same  relative  position  to  a  median 
plane  ;   third,  that  the  surfaces  simply  meet. 

TrjZ-  Reqitireinents  for  Plane  Intersections. — Two  equal  cylinders  of  revo- 
lution— that  is,  cylinders  whose  parallels  are  identical — are  collinearly 
disposed  when  their  axes  are  either  parallel  or  intersect,  thus  affording 
the  longitudinal  or  perimetrical  intersections  respectively. 

Two  equal  prisms  are  similarly  disposed  when,  their  axes  being  paral- 
lel or  made  to  intersect,  a  longitudinal  plane  passing  through  these 
axes  divides  the  surfaces  identically.  The  intersection  is  some  polygon 
which  can  be  cut  on  both  surfaces. 

Two  equal  cones — that  is,  cones  whose  parallels  at  the  same  distances 
from  the  vertices  are  identical — are  collinearly  disposed  when,  their  axes 
being  parallel,  a  plane  cutting  equal  parallels  is  at  the  same  distance 
from  each  vertex ;  or  when  the  axes  are  made  to  intersect  in  a  point 
whose   distance    from    the   vertices  is  the   same.      The  a 

intersection     under     these     conditions     is     either     the  /M    A 

parabola,  hyperbola,  or  ellipse.  /  ^m  ^^ 

Two   equal    pyramids   will    be    collinearly   disposed  /    ^Mv^^ 

when,    in    addition    to    the    conditions    imposed    upon  "^. /__-^^^''j^^i 
the    cones,    the    bases    have    identical    positions    with        \^4^'^0^^ 
reference  to  a  longitudinal  plane  which  passes  through        \  [^/^^^^ 
the  axes.  ^^"^^^^^ 

Any  two  equal  surfaces  whose    homologous  edges  \t-— — "■ 

or  elements  intersect  afford  plane  lines  of  intersection. 

Any  two  surfaces  upon  which  identical  sections  can  r\'^~~-^^^-^'^f\ 

be    cut   may    be   so    disposed    as    to    intersect    in    that  /''/     ; 

section.  kJ^""''^^     "A_y 

274.  Where  surfaces,  such  as  some  of  the  regular 

geometrical   and  surfaces  of  revolution,  can   be   cut  in  /     J\   l\\ 

bi-symmetrical    sections,     an    intersection    may    be    ef-  (/  V   'v,  j 

fected  by  reversing  the  section    on  either  side   of   the  ^^v^    \/^^^ 

cutting  plane.  ^-^ 

275.  By    Analytical    Geometry    it    has   been    demonstrated    that    when 


INTERSECTION^ 


US 


two  cylinders,  or  two  cones,  or  a  cone  and  a  cylinder,  are  made  to  cir- 
cumscribe a  sphere  in  common,  they  intersect  in  a  plane  line  of  inter- 
section. 

276.  Problem. —  To  find  the  line  of  intersection  heticeen  two  equal  eylinders 
Il'/wsc  axes  intersect. 

Let  the  cylinders  be  given  as  in  Fi^.  253,  in  which  the  rcctiUnear 
elements,  having  the  same  position  to  the  coordinate  planes,  give  j)rojcc- 
tions  whicii   make  equal  angles  with   the  ground-line. 

Pass  through  the  two  surfaces  a  series  of  auxiliary  planes  cutting 
rectilinear  elements  upon  each ;  these  planes  are  parallel  to  the  longi- 
tudinal plane  containing  the  axes,  and  give  horizontal  traces  parallel  to 
a"i".  Find  the  points  in  which  each  ])air  of  elements  so  cut  intersect 
each  otlier.  Thus  the  plane  whose  horizontal  trace  is  d"g"  cuts  on  each 
surface  an  element  whose  feet  are  in  the  points  d"  and  g"  respectivelv. 
The  intersection  c'  of  their  vertical  projections  immediately  determines  a 
point  in  the  line  of  intersection  sought,  from  which  the  horizontal  pro- 
jection c"  may  be  found   by  means  of  an  ordinate. 


\ 

,A  \. 

I ' 

\ 

K 

\  '-J 

\{\  i\ 

1     ' 
1 

^^'  1 

In  like  manner  each  auxiliary  plane  determines  a  point  which,  together 
with  the  highest  point  (//'.  //")  found  by  the  plane  passing  through  the 
axes  and  the  lowest  points  found  at  the  intersection  of  the  bases,  fixes 
the  position  of  the  line  of  intersection  sought. 

The  plane  of  the  curve  being  perpendicular  to  //,  c" k"  is  its  hori- 
zontal projection. 

277.  Pkohi.EM.— Ti; ///</  the  line  of  intersection  beticeen  tico  equal  prisms 
;\/'..'.v<    axes  intersect  at  right  angles. 


Ii6 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


Let  the  axes  cd  and  dl  be  given  as  in  Fig.  254,  and  let  a"b"f"  .  .  . 
be  the  horizontal  projection  of  the  base  of  the  vertical  prism. 

Since  the  base  of  the  second  prism  is  perpendicular  to  H,  assume 
any  line,  g" k" ,  perpendicular  to  d" I"  as  its  horizontal  projection.  By 
rabattement  and  counter-rotation  around  the  horizontal  passing  through 
■(/',  I")  determine  the  vertical  projection  g'h'i'k'g',  when  the  homologous 
edges  of  the  surfaces  immediately  intersect  in  points  of  the  line  of  in- 
tersection whose  horizontal  projection  falls  in  the  perimeter  of  the  base 
of  the  vertical  prism. 

278.  Problem. —  To  find  the  line  of  intersection  between  two  eqnal  pyra- 
mids ivJiose  axes  intersect  at  right  angles. 

Let  the  axes  oc  and  oi  be  given  as  in  Fig.  255,  and  let  a"b"d"e"  .  .  . 
be  the  horizontal  projection  of  the  smaller  base  of  the  vertical  pyramid. 

As  the  point  of  intersection  of  the  axes  {o\  0")  must  be  at  the  same 
distance  from  the  vertices  (Art.  273),  make  o"i"  equal  to  c'o',  and  g"k" 
equal   to    b'f.     With   {p\  o")   as   a   centre,    find    the    projections    of    any 

Fig.  ass 


larger  base  parallel  to  H  of  the  vertical  pyramid,  and  by  rotation  of 
the  same  around  a  horizontal  passing  through  {0',  0")  determine  its 
vertical  projection  for  the  second  pyramid.  The  bases  of  the  respective 
pyramids,  being  collinearly  disposed  with  reference  to  the  plane  of  the 
axes,  the  edges  immediately  intersect  in  points  of  the  required  line  of 
intersection. 


INTERSECTIONS. 

279.    Problem. —  To  find  the  line  of  intersection  bettveen  tivo  eqnal  cones. 

Let  a  vertical  cone  be  given  as  in  Fig.  256,  and  let  the  horizontal 
axis  of  the  second   cone  intersect  that  of  the  first  in  the  point  {/',  f"). 

Since  the  axes  of  the  surfaces  intersect  at  the  same  distance  from 
the  vertices  (Art.  273),  make  f"u"  equal  to  f'v',  and  determine  Jt'.  The 
simplest  auxiliary  sections  are  those  cut  by  planes  passing  through  the 
vertices  or  the  line  {v'lc'y  v"u")  which  joins  them.  Such  a  series  of  planes 
cut  rectilinear  elements  on  the  surfaces  and  give  traces  which  contaia 
the  piercing-points  {p' ,  o")  and  (/',  /")  in  common. 

v' 


Thus  the  plane  passing  through  {a\  a")  gives  the  traces  ii"a  and 
ao\  and  cuts  the  vertical  cone  in  the  elements  {a"v'\  a'v)  and  (//":", 
fi'v').  It  likewise  cuts  the  plane  of  the  base  of  the  horizontal  cone  in  a 
line  whose  points  of  intersection  with  that  base  are  the  feet  of  the  ele- 
ments cut  on  the  cone.  One  point  of  the  line  lies  at  the  intersection 
(/',  /")  of  the  horizontal  traces,  the  other  point  (zc",  u^)  at  the  intersec- 
tion of  the  horizontals  {s"u>",  s'w')  and  (</'«'",  o'w'),  found  by  means  of 
the  auxiliary  horizontal  plane  whose  vertical  trace  is  o's'.  Hence,  draw- 
ing the  line  {t'zu',  t"ii<"),  {q,  q")  and  (/-',  r")  are  the  feet  of  the  elements 
sought,  and  (.c'.  .C  )  (''.  <'")  pomts  in  the  line  of  intersection  between 
the  two  cones.  By  a  repetition  ot  this  method  a  sufficient  number  ot 
points  mav  be  found  to  determine  the  entire   line. 


ii8 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


III.      INTERSECTIONS    IN    WARPED    LINES    OF    INTERSECTION. 

280.  For  all  other  positions  not  indicated  in  the  preceding;  twG 
chapters  surfaces  will,  in  general,  intersect  in  lines  which  cani;^t  be 
contained  in  a  single  plane.  As  already  shown,  surface^  w^hich  by 
simply  meeting  would  present  plane  lines  of  intersection  will  when  ex- 
tended give  lines  whose  branches  lie  in  two  distinct  planes. 

Disregarding  these  latter  cases,  which  may  be  considered  as  special 
illustrations  of  the  second  division,  the  class  of  intersections  now  under 
discussion  may  be  grouped  under  the  heading  of  ivarpcd  lines  of  inter, 
section. 

281.  The    warped    lines   of   intersection    may    be   arranged    in   genera! 
classes  according  to  the  nature  of  the  surfaces  which  are 
made  to  intersect. 

(i)  Polyhedral  surfaces  which  meet  always  intersect 
in  rectilinear  intersections,  the  separate 
faces  determining  the  various  plane 
branches  thereof. 

(2)  Polyhedral  and  ruled  surfaces  which 
meet   always  intersect  in  a  broken  curved 
or   a   mixed    line  of   intersection    composed    of    rectilinear 
and    curvilinear   elements. 

(3)  Double-curved  surfaces    intersect   in  double-curved 
lines  of  intersection  of  one  or  two  branches. 

282.  With  a  system  of  polyhedrons  three  classes  of  intersections  may 
be  noted : 

(i)    Where  the  edges  of  one  surface  intersect  those  of  the  other. 

(2)  Where  the  faces  of  one  intersect  the  edges  of  the  other. 

(3)  Where  the  faces  of  one  intersect  the  edges  and  faces  of. the  other. 

283.  With  a  system  of  polyhedral   and   curvilinear  surfaces  the  inter- 
sections  may   be   grouped   according  as   the   second 
surface  is  ruled  or  double-curved. 

(i)  If  ruled,  the  line  of  intersection  may  be 
composed  of  an}-  or  all  the  elements  which  can 
i)e    cut   on   such    a   surface.     Thus   on   the    cone   the 

V  line   may   be  partly   a  right  line, 

!     ^^  when     one     face     of     the     pol)-- 

hedron   coincides  Avith  a  rectilin- 
ear element,  and  partly  any  portion  of  the  curvilinear 
elements. 

(2)    If  double-curved,  the   line   of   intersection   must 
be     wholly    curved,     each     face    of     the     polyhedron 
cutting   a   curve  whose    extremities   are    the   breaking- 
pomts  of  the    line  of  intersection. 

284.  With  a  system  of  two  curvilmear  surfaces  the  lines  of  intersec- 
tion may  be  classified  as  the  surfaces  are  both  ruled,  or  one  ruled  and 
one  double-curved,  or  both  double-curved. 


INTERSECTIONS. 


119 


In  all  of  these  cases  the  line  is  curvilinear,  and  presents  one  or  two 
open  or  closed  branches  sytnmetrically  or  unsymmetricall\-  disposed,  as 
the   relative    position  of   the   two   surfaces  varies. 

285.  Problem. —  To  find  the  line  of  intersection  betzveen  a  right  pyramid 
and  a   right  prism. 

Let  the  surfaces  be  given  as  in  Fig.  257. 

As  the  breaking-points  of  the  line  of  intersection  are  the  piercing- 
points  of  the  edges  on  either  surface,  pass  through  the  axes  and  edges 
a  series  of  auxiliary  cutting  planes.  The  right  elements  cut  on  the  sur- 
faces, and  the  edges  lying  in  the  same  planes,  respectively  intersect  in 
the  breaking-points  required. 

Thus  the  auxiliary   plane    passing  through    the   edge  (/'i'',  f"v")  cuts 


on  the  prism  the  clement  whose  horizontal  pro)ection  falls  in  b'\  thus 
determining  {h\  b")  of  the  line  of  intersection.  In  like  manner  the  plane 
passing  through  the  edge  of  the  prism  whose  horizontal  projection  is  c" 
cuts  on  the  pyramid  the  element  {g'v\  g"r"),  thus  determining  the  point 
ie",  e');  and  as  the  two  surfaces  are  svmmetricallv  disposed  to  each 
other,  the  remaining  points  may  be  found  by  means  of  horizontal  lines 
drawn   through   b'  and  e'. 

286.  Pkor.l.KM. —  To  find  the  line  of  intersection  betzceen  a  right  prism 
and  a  sphere. 

The  intersection  will  be  wholly  curvilinear,  each  face  of  the  prism 
cutting  upon  the  sphere  arcs  of  circles  whose  extremities  are  the  pierc- 
ing-points of  the  edges  upon  that  surface. 

Let  the  surfaces  be  given   as  in   Fig.  25S. 

Pass  through  the  edges  auxiliary   planes  parallel  to   V.  cutting  on  the 


ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 


sphere  circles  parallel  to  that  plane,  and  determine  the  points  of  intersec- 
tion a' ,  a" ,  d',  d" ,  etc.,  between  these  circles  and  the  edges.  Intermediate 
points,  as  (<^',  b")  and  ic' ,  c"),  may  be  found  by  means  of  similar  cutting 
planes,  in  sufficient  number  to  mark  the  entire  line  of  intersection. 

287.   Problem. —  To  find  the  Ime  of  ititerscction  between  two  oblique  prisms. 

Let  the  surfaces  be  given  as  in  Fig.  259. 

The  simplest  auxiliary  planes  are  those  which,  being  parallel  to  the 
axes  of  the  two  prisms,  cut  rectilinear  elements  on  each. 

To  determine  the  position  of  such  a  series  of  planes,  from  any  point 
{a',  a")  of  one  axis  draw  a  parallel  {a'b',  a"b")  to  the  other,  and  through 
their  piercing-points,  <:"  and  b" ,  draw  the  horizontal  trace  of  the  plane 
which  contains  them.  The  horizontal  traces  of  all  the  auxiliary  cutting 
planes  are  parallel  to  c"b" . 

Thus  the  plane  passing  through  the  edge  whose  foot  is  d"  gives  a 
horizontal  trace  d"g"  "Sindi  cuts  the  second  prism  in  the  edge  whose  foot 
is  g"  \  the  point  {p' ,  o")  common  to  the  two  edges  is  a  point  in  the  line 
of  intersection  sought. 


iKig.  359 


Again,  the  plane  passing  through  the  edge  whose  foot  is  e"  cuts  on 
the  second  prism  the  elements  whose  feet  are  h"  and  k"  •  the  points 
{m',  m")  and  {n\  n")  in  which  the  edge  crosses  these  elements  are  likewise 
points  in  the  line  sought.     The  repetition  of  this  method  for  the  other 


IN  TERSECTIONS. 


edges  of  both  surfaces  determines  the  breaking-points  necessary  to  de- 
scribe the  entire   line  of  intersection. 

288.  Problem. —  To  find  the  /inc  of  intersectioji  bctzL'ccn  a  cylinder  and 
a  cone. 

Let  the  surfaces  be  given  as  in   Fig.  260. 

Through   the  two  surfaces  pass  a  series  of  secant  planes  cutting  right 

I) 


elements  on  each.     To  effect   this  lead    through   the  vertex  (b\  b")  of  the 
cone    a    line  (/'  Ti-'.  b"tc")  jxu-allcl    to    the    axis  of  the    cylinder,  and  find  its 


122  ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 

horizontal  piercing-point,  tv" .  Any  secant  plane  passing  through  this 
line  will  cut  the  desired  elements. 

Assume,  then,  the  horizontal  traces  of  such  a  series  of  planes,  all  of 
which  pass  through  the  piercing-point  w" ,  and  determine  the  elements 
in  which  they  cut  the  respective  surfaces ;  their  intersections  are  points 
in  the  line  of  intersection  sought.  Thus,  the  plane  L  cuts  the  cylinder 
in  two  elements  whose  feet  are  a"  and  k" ,  and  the  cone  in  two  elements 
whose  feet  are  o"  and  p" ;  the  points  (/,  r")  {s\  s")  (/',  /")  and  (?/',  ti") 
in  which  they  intersect  are  points  of  the  line  of  intersection.  In  a  similar 
way  a  sufficient  number  of  points  may  be  found  to  determine  the  entire 
line  of  intersection. 

Note. — In  determining  the  respective  portions  which  are  either  seen 
or  concealed  on  the  combined  projections  of  the  intersecting  surfaces  it 
is  to  be  observed  that — 

(i)    The  outline  of  the  united  projections  is  always  visible. 

(2)  The  line  of  intersection  being  common  to  both  surfaces,  the  ver- 
tical projection  of  that  line  is  visible  only  so  far  as  it  lies  on  the  front 
portion  of  either  surface. 

(3)  The  horizontal  projection  of  that  line  is  visible  onl}^  so  far  as  it 
falls  on  the  iippcr  portion  of  either  surface. 

(4)  In  passing  from  the  part  seen  to  that  which  is  hidden,  the  visible 
limits  always  terminate  at  the  apparent  contour  of  either  surface. 

289.  Problem. —  To  find  the  intersection  betzvccti  a  cylinder  and  a  sphere, 
the  sphere  tangent  interiorly  to  the  cylinder. 

Let  the  cylinder  be  given  as  in  Fig.  261,  and  let  the  centre  of  the 
spnere  lie  in  the  meridian  plane  passing  through  the  element  whose  foot 
is  {a'\  a'). 

Determine  the  centre  of  the  sphere  by  rabattement  of  the  meridian 
plane  around  HL,  carrying  with  it  the  element  and  the  meridian  sec- 
tion of  the  sphere  tangent  to  it.  By  counter-rotation  the  rabattement 
of  the  centre  c-/^  the  tangent  point  //'  and  the  highest  point  ///'  of  the 
line  of  intersection  fall  respectively  in  c" ,  t"  and  //'.  From  e"  as  a 
centre  describe  the  horizontal  projection  of  the  sphere,  and  from  c' — 
determined  by  measuring  the  distance  c"c^'  above  H — the  vertical  pro- 
jection of  the  same. 

The  parallels  of  the  two  surfaces  being  the  simplest  constructive  ele- 
ments, pass  between  the  highest  and  lowest  points,  h'  and  the  bases, 
respectively,  a  series  of  horizontal  cutting  planes ;  the  points  in  which 
the  sections  intersect  are  points  in  the  line  of  intersection  sought. 

Thus  the  plane  M  cuts  on  the  sphere  the  circle  whose  radius  is  {i'b' , 
i"b"),  and  on  the  cylinder  the  circle  whose  centre  is  (/,  e"),  which  in- 
tersect each  other  in  the  points  {d',  d")  and  ig',  g").  Similar  con- 
structions for  the  other  secant  planes  mark  a  sufficient  number  of  points 
to  determine  the  entire  curve  of  intersection.  Should  the  sphere  be 
entire,  the  curve  thus  determined  is  composed  of  two  closed  branches 
which  cross  at  the  point  of  tangency. 


IiVTERSECT/ONS. 


123 


290.  Prohi.F.M. —  To  find  the  i)iti'rsiction  bi't'iK'Ccn  two  surfaces  of  rrt'o/n- 
iion  ti'hosc  axes  lie  in  a  couunon  meridian  plane. 

Let  the  two  surfaces  be  given  as  in  Fig.  262,  the  axes  intersecting 
in  the  point  (/',  /"). 


Tn  such  a  case  plane  auxiHarv  sections  do  not  present  simple  con- 
structive elements:  but  if  the  point  (/',  /")  be  assumed  as  the  centre  of 
a  series  of  auxiliarv  spheres,  then  each  surface  will  be  intersected  in  a 
circle  the  planes  of   which  are  perpendicular  to  the  respective  axes.     As 


124 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


the  circles  so  cut  are    common  to  the  spheres,   they  intersect   in   points 
of  the  line  of  intersection  sought. 

Thus  the  first  auxiliary  sphere  intersects  the  ellipsoid  in  the  circle 
whose  vertical  projection  is  a'b' ,  and  the  paraboloid  in  the  circle  whose 
vertical  projection  is  c'd' ;  the  two  points  in  which  these  circles  inter- 
sect coincide  in  c'  and  are  points  in  the  line  of  intersection.  The  points 
(/'>  f")  {s' 1  s")  found  by  means  of  additional  spheres,  together  with 
the  highest  point  (//,  //")  and  the  lowest  point  (/',  I"),  are  sufficient  to 
determine  the  entire  line. 


291.  Problem. —  To  find  the  intersection  between  two  surfaces  of  revolu- 
tion whose  axes  do  not  lie  in  the  same  plane. 

Let  the  surfaces  be  given  as  in  Fig.  263,  the  axes  of  both  surfaces 
being  parallel  to   V. 

In  this  position  it  is  possible  neither  to  employ  secant  planes  to  cut 
the  surfaces  in  parallels,  nor  auxiliary  spheres  to  cut  them  in  circles. 
The  simplest  constructive  method  which  remains  is  the  use  of  horizontal 
planes  giving  on  the  first  surface  parallels,  and  on  the  second  sections 
whose  horizontal   projections   must  be  determined   by   points. 

Thus,  let  VL  be  the  vertical  trace  of  such  a  secant  plane,  cutting 
the  first  surface  in  a  circle  ia'b' ,  a"  .  .  .  b"),  and  the  second  in  a  section. 


IN  TERSE  C  TIONS.  1 2  5 

-whose   vertical   projection    is   the    line   c'd'  and   whose    horizontal  projec- 
tion may  be  found  by  the  following  construction : 

Pass    through    the    surface    a    series    of    planes    perpendicular  to    the 

axis  (//',  e"f").     Let   x'y'  be   the   vertical    projection  of   a  circle  cut    by 
such  a  plane;    then   the   point  z'  in   which   it  intersects  c'd'  is  a  point  of 

that  auxiliary  section,   and    by  the   rabattemcnt  of   the    circle   x'y'  around 

FiS»S03 


the  honzwntal  projected  vertically  in  z  the  horizontal  projection  z"  may 
be  determined.  In  a  similar  way  any  number  of  points  of  the  auxiliary 
section  {c'z'd',  c"z"d")  may  be  obtained.  The  curve  thus  found,  and  the 
circle  whose  vertical  projection  is  a'b',  lying  in  the  same  plane,  inter- 
sect in  two  points  (w',  m")  and  (//',  //")  of  the  line  of  intersection  sought. 
By  repetition  of  this  method  a  series  of  points  may  be  found  through 
which   the  entire   line  of  intersection   hiklmn   passes. 


1 2D 


ELEMENTS   OF  DESCRIPTIVE   GEOMETRY. 


CHAPTER  XIII. 

TANGENTS    AND    NORMALS. 
I.     GENERAL   CONSIDERATIONS. 

292.  Any  right  line  lying  in  the  plane  of  a  curve  and  intersecting^ 
it  in  two  points  a  and  b,  which  are  not  consecutive  (Fig.  264)  or  are 
separated  by  a  finite  distance,  is  a  secant  to  that  curve.  Should  such 
a  Une  be  turned  around  «  as  a  pivot,  the  point  b  gradually  shortens  its 
distance  from  a  until  it  reaches  a  position  in  which  it  becomes  consec- 


Fig.  204 


Eig.2e5 


utive  with  it.  In  this  position  ab'"  no  longer  intersects  the  curve,  and 
is  said  to  be  tangent  to  it.  Should  the  motion  be  continued,  the  points 
again  begin  to  separate  and  the  line  becomes  a  secant. 

Hence  the  tangent  may  be  defined  to  be  the  limit  of  a  secant  drawn 
from  one  point  of  a  curve  to  another  point  of  that  curve  at  an  in- 
finitely small  distance   apart. 

The  point  a  is  termed  the  point  of  contact,  and  the  line  ac,  drawn  at 
right  angles  to  the   tangent  ab'",   the   normal  to  the  curve  at  that  point. 

293.  In  a  like  manner  two  curves  may  be  tangent  to  each  other 
(Fig.  265)  when,  lying  in  the  same  plane,  their  points  of  intersection 
become  consecutive,  or  when  a  line  tangent  to  one  is  at  the  same  time 
tangent  to  the  other  at  a  common  point.  In  such  a  case  the  normal  ac 
is  likewise  common  to  both  curves,  the  point  a  in  which  it  intersects 
the  tangent  being  termed  the  point  of  incidence. 

294.  The  normal  plane  to  any  point  of  a  curve  is  the  plane  led  through 
this  point  at  right  angles  to  the  tangent  line. 

295.  A  point  of  inflexion  upon  a  branch  of  any  curve  is  one  in  which 


TANGENTS  AND   NORMALS. 


127 


the    normal  plane  and  every  tanj^ent  plane  at  this  point  with    one  exccp- 
tion  crosses  it  fFig.  266,    1 1. 

A  cusfy  point  is  that  in  which  two  branches  of  a  curve  have  a  com- 
mon tanjj;-ent  and  are  located  on  the  same  side  of  the  normal  plane  (Fig. 
266,  2,  3). 


Kig.  sen 


The  angle  of  two  curves  is  the  angle  measured  by  their  tangents  at  a 
common     point.      The    curves    are    tangent    when     this 
angle   is   a   minimum,   and   normal  when   it  is   right. 

296.  A  rectilinear  tangent  to  a  surface  is  a  line 
that  touches  any  curve  on  the  surface  cut  by  a  secant 
plane   which   passes   through   that   line. 

297.  A  ])lane  is  tangent  to  a  sur- 
face when  it  has  one  point  in  common 
with  it,  or  when,  if  any  secant  plane 
be  passed  through  that  point,  the  right 
line  cut  on  the  plane  and  the  curve 
cut  on  the  surface  are   tangent  to  each   other. 

Hence  the  tangent  plane  to  a  surface  may  be  con- 
sidered as  the  geometrical  locus  of  the  tangents  at  the 
point  of  contact. 
A  normal  to  a  surface  is  the  line  drawi^  at  right  angles  to  the 
tangent  plane  through  the  point  of  contact,  ami  any  plane  passing 
tiirough   a   normal   is  a   normal  plane. 

Oiilv    one    normal    can     be     drawn    at    anv    j)oint    of 
an    indehnite    number   of    normal    planes,    all    of    which 
])ass    through    or   intersect   one    another   in   the   normal 
line. 

299.  Two  surfaces  arc  tangent  to  each  other  when 
a  plane  is  tangent  to  both  at  the  common  element  of 
contact,  or  when,  if  anv  secant  j)lane  be  passed 
through  both,  the  line  cut  on  the  plane  is  tangent  to 
the  sections  cut  on  the  surfaces  at  a  common  point 
of  contact. 

300.  The   angle  of  t-wo  surfaces    at    any    point    is    the    angle    of    their 
tangent   planes,  or  of  their  normals  at  that  point.     When  the  angle  is  a 


incidence,    but 


128  ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 

minimum  the  surfaces  are  tangent;    when  right  they  are  normals  to  each 
other. 

301.  From  the  foregoing  it  follows  that— 

(i)  Every  plane  passed  through  the  point  of  contact  on  a  surface 
cuts  a  section  on  the  surface  and  a  line  on  the  tangent  plane  which 
are  tangent  to  each   other  at  that  point. 

(2)  The  number  of  secant  planes  which  can  be  passed  through  the 
point  is  unlimited,  giving  in  every  case  a  right  line  on  the  plane  and 
a  section  on  the  surface  which  are  tangent  to  each  other. 

(3)  All  the  sections  thus  cut  have  one  point  in  common — the  point 
of  contact. 

(4)  All  lines  drawn  through  the  point  of  contact  tangent  to  the  sur- 
face are  lines  of  the  tangent  plane. 

(5)  Conversely,  all  lines  in  the  tangent  plane  which  are  tangent  to 
the  surface  pass  through  the  point  of  contact. 

(6)  Tangents  and   normals  are   conjugate. 

(7)  Lines  tangent  to  each  other  in  space  give  projections  which  are 
tangent. 

302.  As  the  tangent  plane  is  the  locus  of  all  tangent  lines  at  any 
given  point,  any   two  tangents  will  determine  that  plane. 

A  normal  may  be  determined  by  the  tangent  plane  since,  being  per- 
pendicular to  that  plane,  its  projections  are  perpendicular  to  the  respect- 
ive traces. 

11.   TANGENTS  AND  NORMALS  TO  SURFACES. 

303.  Rtiled  Surfaces. — With  single-curved  surfaces  the  tangent  plane 
passes  through  or  contains  a  right  element — tJie  line 
of  contact.  If  such  a  surface  and  its  tangent  plane  be 
cut  by  any  secant  plane,  the  section  on  the  surface 
and  the  line  on  the  plane  will  be  tangent  to  each 
other  in   a  point  of  the   line    of  contact. 

Hence  the  tangent  plane  gives  a  trace  tangent  to 
the  base,  cut  by  either  plane  of  projection,  and  at  the 
foot  of  the   line   of  contact. 

304.  A  single  point  on  a  ruled  surface  determines 
a  tangency,  since  it  fixes  the  position  of  the  line  of  contact  and  a 
second   tangent  line   of  the  required  plane. 

305.  Through  a  point  in  space  a  tangent  plane  cannot  always  be 
passed.  The  conditions  demanded  in  such  a  case  will  be  satisfied  when, 
with  cylindrical  surfaces,  the  point  lies  beyond  and  towards  the  convex 
side,  and,  with  conical  surfaces  when  the  line  joining  the  point  with 
the  vertex  passes  clear  of  the  nappes. 

306.  Through  a  line  in  space  a  tangent  plane  can  be  passed  when 
the  line  is  either  parallel  to  an  element  or  is  tangent  to  any  section 
cut   on    the    surface.     Under   these    conditions    there   will    be  one   or  two 


TANGENTS  AND   NORMALS.  I29 

planes,  as  the  line  is  tangent  to  a  section  or  passes  through  the  vertex 
of  a  cone,  or  is  tangent  to  a  section  or  parallel  to  the  axis  of  a  cylin- 
der. 

307.  Parallel  to  a  line  in  space  two  tangent  jjlanes  can.  in  general, 
be  passed,  but  parallel  to  a  plane  only  <jne  can  be  led  to  the  cone  and 
two  to  the  cylinder. 

308.  In  the  ruled  surfaces  the  normals  ha\'e  their  points  of  incidence 
in  the  line  of  contact,  being  for  any  one  element  parallel  to  one  another 
and,  hence,  lying  in  the  same  plane — the  normal  plane. 

309.  Doublc-curvi'd  surfaces  have  but  one  point  of  tangency  in  com- 
mon with  a  plane. 

Through  a  point  in  space  an  unlimited  number  of  tangent  planes 
may  be  passed  to  a  double-curved  surface,  inasmuch  as  an  infinite  num- 
ber of  tangent  Imes  can  be  drawn  which,  instead  of  forming  a  tangent 
plane,  as  in  the  cases  of  the  ruled  surfaces,  determine  a  conic  surface, 
through  each  element  ol  which  a  tangent  plane  may  be  led. 

310.  Parallel  to  a  line  in  space  an  unlimited  number  of  tangent  lines 
and  planes  can  be  passed  to  the  surface,  the  lines  determining  a  cylin- 
drical surface  and  forming  with  their  points  of  contact  a  continuous  line. 

Parallel  to  a   i)lane  in  space  but  two  tangent  planes  can  be  passed. 

311.  In  the  case  of  the  sphere  all  tangent  lines  and  planes  are  per- 
pendicular to  the  respective  radii  at  the  points  of  contact;  hence  the 
traces  of  such  a  plane  are  always  perpendicular  to  the  {projections  of 
the  radii  passing  through  these  points. 

As  all  the  radii  are  normals,  it  follows  that  all  normal  planes  contain 
the  centre  and  cut  great  circles  on  the  surface. 

With  the  great  circles  the  normal  surface  determined  by  the  infinite 
series  of  normal  lines  is  a  plane,  but  with  the  smaller  circles  it  is  a  com 
whose  vertex  is  at  the  centre. 

312.  Surfaces  of  Rotation. — Through  any  point  on  such  a  surface  a 
parallel  and  a  meridian  section  can  be  cut,  the  rectilinear  tangents  to 
^vhich   fix  the  position  of  a  tangent  plane. 

313.  Tangents  to  parallels  determine  parallel  planes,  but  tangents  to 
meridians  determine  either  a  cylindrical  or  conical  surface:  cylindrical 
when  the  tangents  are  parallel  to  the  axis,  and  conical  when  inclined, 
since,  lying  in  the  plane  of  the  meridian,  they  necessarily  intersect  the 
axis  111  a  common  point. 

314.  The  tangent  plane  is  perpendicular  to  its  conjugate  meridian 
plane,  and  its  inclination  to  the  axis  is  measured  by  the  tangent  line  to 
the  meridian   at  the  point  of  contact. 

315.  The  normal  surface  to  a  meridian  is  a  plane,  but  to  a  parallel 
it  is  a  cone  whose  vertex  lies  in  the  axis. 

316.  In  connection  with  the  point,  line,  or  plane  in  space  the  same 
general  conditions  prevail  as  with  the  double-curyed  surfaces. 

317.  Prohlkm. —  To  pass  a  line  tangent  to  a  section  of  the  cylinder,  the 
point  of  contact  being  given. 


T50  ELEMENTS  OF  DESCRIPTIVE   GEOMETRY. 

Let  ib' ,  V)  be  the  assumed  point  of  contact  (Fig.  240). 

As  the  tangent  line  is  common  both  to  the  secant  and  tangent  planes^ 
it  is  their  line  of  intersection.  Hence  at  the  foot  b"  of  the  element  of 
contact  draw  the  horizontal  trace  HL  of  the  tangent  plane ;  the  point 
k"  in  which  it  intersects  HM  is  a  point  of  the  required  tangent,  and  its- 
vertical  projection  k'  a  point  of  the  vertical  projection  k'b'  of  that  line. 

318.  Problem. —  To  pass  a  line  tangent  to  a  section  of  a  cone,  the  point 
of  contact  being  given. 

Let  (//,  //')  be  the  assumed  point  of  contact  (Fig.  241). 

As  in  the  preceding  case  draw  the  trace  HL  of  the  tangent  plane, 
and  find  its  intersection  p"  with  HM.  The  point  $0  found  determines 
the  tangent  p" h" ,  whose  vertical  projection  coincides  with  the  trace  of 
the  secant  plane. 

To  determine  the  tangent  in  its  revolved  position,  find  the  vertical 
piercing-point  o'  of  the  tangent  line — since  it  remains  fixed  during  the 
rabattement  of  the  secant  plane — and  the  line  o'h^'  is  the  tangent  sought. 

319.  Problem. —  To  pass  a  line  tangent  to  a  section  on  a  cylinder. 
Let  ix' ,  x")  be  the  assumed   point  of  contact  (Fig.  246). 

At  the  foot  d"  of  the  element  of  contact  draw  the  horizontal  trace 
HL  of  the  tangent  plane,  and  find  the  horizontal  piercing-point  //'  of 
the  line  of  intersection  between  this  plane  and  the  secant  plane  M ,  the 
point  so  determined,  in  connection  with  the  assumed  point,  fixes  the  posi- 
tion of  the  required  tangent  {Ji' x' ,  h'^x"). 

As  the  secant  plane  is  brought  into  the  horizontal  plane,  the  piercing- 
point  //'  remains  fixed  in  its.  position ;  hence  it  will  be  the  second  point 
necessarv  to  draw  the  revolved  tangent  line  h" x^' . 

320.  Problem. —  To  find  a  tangent  to  any  plane  curve  not  subject  to  geo- 
metrical definition. 

Let  A  .  .  C  .  .  F  hQ  any  curve  drawn  at  will,  and  let  it  be  required  to 
draw  a  tangent  at  the  point  D. 


Take  any  number  of  points,  as  A,  B,  C,  etc.,  on  either  side  of  D  and 
draw  a   series  of   secants    throug^h    them    and   L>.     Describe   from  Z>  as  a 


TANGENTS  AND   NORMALS. 


131 


centre  an  arc  of  any  radius,  cutting  the  secants  in  points  a,  b,  c,  etc., 
and  lay  off  on  the  secants  distances  equal  to  the  respective  chords  of 
the  orit^inal  curve.  Thus,  make  in  =  AD,  bz  =  BD,  c^  =  CD,  c^  —  ED, 
etc.;  the  curve  traced  through  the  points  i,  2,  3,  etc.,  intersects  the 
circle  in  the  point  d  of  the  required  tangent,  since  for  that  line  the 
chord   is  reduced   to  a  minimum. 


III.  PROBLEMS.   TANGENTS  TO  RULED  SURFACES. 

321.  Probleal — To  pass  a  plane  tangent  to  a  right  cylinder  at  any  point 
of  a  section  and  determine  the  tangent  line  to  that  section. 

Let  the  cylinder  be  given  as  in  Fig.  267,  and  let  {J'M,  HM)  be  the 
traces  of  the  secant  plane,  and  {a' ,  a")  the  given  point. 

The  tangent  plane  contains  the  element  of  which  {a' ,  a")  is  a  point ; 
hence  its  horizontal  trace  HL  is  tangent  to  the  base  of  the  cylinder  at 
the  foot  a" ,  and  its  vertical  trace    VL  is  perpendicular  to  GL. 

The    tangent   to   the    section    is   the    line   of   intersection   between    the 


tangent  and  secant  planes  lArt.  297);  hence  its  projections  coincide  re- 
spectively with  VM  and  HL.  When  the  plane  J/  is  brought  by  rabatte- 
inent  around  VM,  the  piercing-point  (/'.  /")  of  the  line  of  intersection  or 
tangent  remains  fixed  ;    hence  the   line  p'a,'  is  the  tangent  required. 

322.  Problenl — Through  a  given  point  in  space  to  pass  a  plane  tangent 
to  a  cylinder. 

Let  the  cvlinder  be  given  as  in  Fig.  2CS,  and  let  {p',p")  be  the  given 
point. 

As  the  tangent  plane  must  pass  through  an  element  of  the  surface, 
and  hence  be  parallel  to  the  axis,  the  line  ip'o',  p"o")  led  through  the 
given  point  will  be  a  line  of  the  tangent  jilane,  and  its  piercing-point 
o"   a    point  of   the  trace  HM. 


132 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


The  vertical  piercing-point  n  is  a  point  of  the  vertical  trace  VM\ 
but  as  the  traces  intersect  beyond  the  limits  of  the  drawing,  an  addi- 
tional  point  must   be  found.     To  effect  this,  pass  through  any  point  {b\ 


ji^g.^.6a 


b")  of  the  line  of  contact  {b'b' ,  b"b")  a  horizontal  {h'm\  b"m")  of  the  plane 
M,  and  find  its  vertical  piercing-point  in' ;  the  line  joining  n'  and  in'  is 
the  trace  required. 

A  second  tangent  plane  L  may  be  determined  by  a  similar  construction. 

323.  Problem. —  To  pass  a  plane  tangent  to  a  cylinder  and  parallel  to  a 
^ivcn  line. 

Eig.  369 


Let   the    cylinder  be  given  as  in   Fig.  269,  and   let  {a'b' ,  a"b")  he  the 
a:iven  Une. 


TANGEN'J-^  AXD  XOKMALS. 


'35 


IVom  any  j)()int  {c\  c"^  of  the  line  let  fall  a  parallel  {c'd' ,  c"d")  to 
the  axis  of  the  cylinder,  and  determine  the  horizontal  trace  HL  of  the 
plane   which  contains  the  line  so  found  and  the  given  line  (Art.  85). 

As  any   plane  tanj^-ent  to  the  cylinder    must    be    parallel    to    the    axis, 
and  as  by  the  conditions  of  the  problem  it  must   be  likewise   parallel  to 
the  given   line,  the  plane   J/,   whose    hi^rizontal    trace  HM   is   parallel    to 
}IL   and    tangent   to   the    base,  satisfies   both   requirements,    and    hence    i 
the  horizontal  trace  of  the  required   plane. 

To  determine  the  vertical  trace  assume  any  point  (^',  ij" )  of  the  cle- 
ment of  contact,  and  draw  through  it  a  line  parallel  to  HM\  the  line 
so  drawn  is  a  horizontal,  and  its  piercing-point  is' ,  s")  in  connection 
with  the  trace  HM  fixes  the  position  of  the  vertical  trace   VM. 

By  similar  methods  of  construction  a  second  plane  N  may  be  deter- 
mined. 

324.  Problem. — Given  one  projection  of  a  point  on  the  surface  of  a  cone, 
to  pass  through  that  point  a  plane  tangent  to  the  cone. 

Let  the  cone  be  given  as  in  Fig.  270,  and  let  a'  be  the  vertical  pro- 
jection of  the  given  point. 

Pass  through  the  point  a  right  element  {b'p',  b"p")  of  the  surface, 
and  determine  a"  by  means  of  the  ordinate.  The  tangent  plane  must 
contain  this  element,  and  hence  its  horizontal  trace  IIL  must  pass  through 
the  piercing-point  /"  and  be  tangent  to  the  base. 

To   find    VL,    draw    through    any  point   of  the    clement  of   contact,  as 


{a',   a"),   a    horizontal   {a'o',   a"o")   of    the    plane    L 
Ki-.i2ri 


its    vertical    piercing- 


point  (/.  in  connection  with  the  horizontal  trace  HL,  determines  the  trace 
sought. 

325.  Problem. —  To  pass  through  a  point  in  space  a  p/anc  tangent  to  a 
cone. 

Let  the  cone  be  given  as  in  Fig.  2j\.  and  let  {a  ,  a'  )  be  the  given 
ponit. 


134 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


As  every  plane  tangent  to  a  cone  passes  through  the  vertex,  the  Hne 
{a'v ,  a"v")  joining  the  given  point  and  the  vertex  must  be  a  line  of 
the  tangent  plane,  and  its  horizontal  piercing-point  /"  a  point  of  the 
horizontal  trace  HL  or  HM.  The  vertical  trace  VL  or  VM  passes 
through  the  vertical  piercing-point  o'  of  this  same  line. 

A  second  point  of  the  vertical  trace  may  likewise  be  determined  by 
means  of  a  horizontal  of  the  tangent  plane  or  by  the  element  of  contact. 

326.  Problem. —  To  pass  a  plane  tangent  to  a  cone  and  parallel  to  a  givoi 
right  line. 

Let  the  cone  be  given  as  in  Fig.  272,  and  let  R  be  the  given  line. 

The  line  i^o'p' ,  v"p")  led  through  the  vertex  and  parallel  to  the  given 
line  must  be  a  line  of  the  tangent  plane,  since  every  tangent  plane 
passes  through  the  vertex  and,  by  the  conditions  of  the  problem,  must 
be  parallel  to  the   given    line. 

Find  the  piercing-points  of  the  line  {z'p' ,  v"p")  and  proceed  as  in  the 
preceding  case. 

IV.    SURFACES    OF    REVOLUTION. 

327.  Problem. —  ThrongJi  a  point  on  a  surface  of  revolution  to  pass  a 
plane  tange?it. 


Let  the  surface  be  given  as  in  Fig.  273,  and  let  a"  be  the  horizontal 
projection  of  the  point. 

As  the  tangent  plane  is  perpendicular  to  the  meridian  plane  of  a 
surface  at  the  point  of  incidence  of  the  normal,  and  passes  through  the 
tangent  to  the  meridian  section  at  that  point,  through  ia' ,  a")  pass  a 
secant  plane  cutting  the  meridian,  of  which  c"d"  is  the  horizontal 
projection.  Revolve  this  plane  around  the  axis  of  the  surface  until  it 
assumes  a  position  parallel  to  V,  carrying  with  it  the  point,  meridian, 
tangent    and    normal ;    these    constructive    elements  are    respectively    pro- 


TANGENTS  AND  NORMALS. 


135 


jected  at  {a,',  a,"),  {d,'c\\  d^'c^'),  («,V/,  «,'V,"j  and  («,'/',  a,"/").  By 
counter-rotation  the  foot  of  the  tangent  line  falls  at  c" ,  and  the  normal 
coincides  with  the  horizontal  trace  of  the  secant  plane.  As  the  tangent 
plane  is  perpendicular  to  the  normal  af,  ML  drawn  thnjugh  c'\  and 
perpendicular  to  that  line,  is  the  horizontal  trace  sought. 

The  vertical  trace  VL  is  found  by  leading  through  the  ground-point 
a  perpendicular  to  the  vertical  projection  fa'   of  the  normal. 

328.  Problem. —  Through  a  given  point  in  space  to  pass  a  plane  tangent 
to  a  sphere  at  a  given  meridian. 

Let  the  sphere  be  given  as  in  Fig.  274,  and  let  {a\  a" )  be  the  point, 
and  HL  the  trace  of  the  meridian  plane. 

Any  plane  tangent  to  a  surface  of  revolution  is  tangent  to  every 
section  cut  on  that  surface  by  secant  planes  which  pass  through  the  point 
of  contact.     In  the  present  case  one  of  the  sections  is  the  meridian,  and 


a  second  may  tx^  assumed  to  be  a  parallel  or  circle  passing  through  that 
same  point.  The  rectilinear  tangents  to  these  curves  are  at  right  angles 
to  each  other  and  fix  the  position  of  the  tangent  plane. 

Hence,  through  the  given  point  {a\  a'')  let  fall  a  perpendicular  {a'b\ 
a"b")  upon  the  meridian  plane,  A.  and  find  its  |iicrcing-point  {b\  b") 
thereon.  Through  this  latter  point  lead  tangents  t(»  the  meridian;  these 
tangents,  together  with  the  line  {a'b' ,  a"b' )  which  is  parallel  to  the  tangent 
of  the  circle,  serve  to  fix  the  position  of  the  tangent  planes  required. 

To  determine  the  tangents  revolve  the  meridian  plane  around  the 
vertical  axis  of  the  sphere  until  it  assumes  a  position  parallel  to  l\ 
when   b"  falls   at   b^"  and    the    vertical    projection    of   the    meridian   coin- 


136 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


cides  with  the  apparent  contour  of  the  surface.  As  the  line  ah  is  horizon- 
tal, its  vertical  projection  a'b'  is  parallel  to  GL  and  contains  the  point  b^ . 
Hence  b^c^  and  b^d^  are  the  revolved  tangents  sought,  and  (t/,  t/'j  (///,  c//') 
their  points  of  contact.  By  counter-rotation  (f/,  <:/')  falls  at  (<:',  r"),  (///,  d^') 
at  {d' ,  d"),  and  the  feet  of  the  tangents  fall  at  {e" ,  e')  and  (/",  /'). 

As  the  tangent  plane  is  perpendicular  to  the  normal  at  the  point  of 
contact,  the  lines  HN  and  HAI  drawn  through  c"  and  f"  respectively, 
and  perpendicular  to  the  horizontal  projections  o"c"  and  o"d''  of  the 
normals,  are  the  horizontal  traces  sought.  The  vertical  traces  VN  and 
HN  are  found  by  leading  through  the  ground-points  perpendiculars  ta 
the  respective  vertical  projections  o'c'  and  o'd'  of  the  normals. 

329.  Problem — To  pass  a  plane  tangent  to  a  surface  of  revolution  and 
parallel  to  a  given  plane. 

Let  the  surface  be  given  as  in  Fig.  275,  and  let  L  be  the  given  plane. 


Any  plane  tangent  to  the  surface  is  perpendicular  to  the  plane  of 
the  meridian  passing  through  the  point  of  contact;  hence  the  given 
plane  L  and  the  required  plane  being  parallel  are  both  perpendicular 
to  a  common  meridian  plane,  which,  according  to  Geometry,  intersects 
them  in  parallel  lines.  If,  then,  the  plane  M  be  passed  through  the 
axis  of  the  surface  perpendicularly  to  the  plane  Z,  it  will  cut  the  re- 
quired meridian  on  that  surface  and  a  line  on  the  plane,  to  which  line 
the  tangent  to  the  meridian  will  be  parallel. 

To  determine  this  tangent  find  the  line  of  intersection  between  the 
planes  L  and  M\  {a'b',  a"b")  are  the  projections  required.  Now  revolve 
the  plane  M  around  the  axis  of  the  surface  until  it  assumes  a  position 
parallel  to  F,  carrying  with  it  the  line  {a'b' ,  a"b")  and  the  meridian 
section;  the  former  falls  at  ia^b',  a."b^"),  and  the  latter  coincides  in 
vertical  projection  with  the  apparent  cont  )ur  of  the  surface. 


TANGENTS  AND  NORMALS. 


137 


It  is  evident  that  the  lines  r,V,'  and  ^///,',  drawn  parallel  to  a^b^,  are 
the  vertical  projections  of  the  tan^^ents  in  the  revolved  position.  By 
counter-rotation  they  are  projected  in  the  horizontal  trace,  IIM,  of  the 
meridian  plane,  their  feet  falling  ni  the  points  /"  and  c"  respectively. 
The  horizontal  traces  of  the  required  planes  pass  through  these  points 
and  are  parallel  to  HL  of  the  given  plane :  the  vertical  traces  pass 
through  the  ground  p<}ints  and   are  parallel  to    \'L. 

A  verification  of  the  solution  will  be  found  in  the'  fact  that  the  traces 
are  perpendicular  to  the  respective  normals. 

330.  Problem. —  To  pass  a  plane  parallel  to  a  given  line  and  tangent  to 
a  surface  of  revolution  at  a  given  ineridiaji. 

Let  the  surface  be  given  as  in  Fig.  276,  and  let  {a'b',  a"b")  be  the 
given  line,  and  HiM  the  horizontal  trace  of  the  meridian  plane.    . 


Fig.aro 


The  tangent  plane  contains  the  two  tangents,  one  to  the  meridian 
section  and  one  to  the  parallel  or  circle  at  the  point  of  contact. 

Hence,  from  any  point  {a',  a")  of  the  given  line  draw  a  parallel  to 
the  tangent  to  the  circle — that  is,  a  perpendicular  to  the  meridian  plane  ; 
the  plane  L  which  passes  through  these  two  lines  is  parallel  to  the 
tangent  plane,  and  cuts  the  plane  of  the  meridian  in  a  line  parallel 
to  the  tangent  line.  The  problem  is.  then,  resolved  into  the  preceding 
case.  Thus,  the  plane  L  intersects  the  meridian  plane  J/  in  the  line 
ie'd',  e"d"),  to  which  the  tangent  to  the  meridian  must  be  drawn  parallel. 
To  effect  this  construction,  revolve  the  meridian  plane  around  the  axis 
of  the  surface  until  it  assumes  a  position  parallel  to  V,  and  determine 
the  tangent  lines  as  in  Fig.  275.  By  counter-rotation,  the  traces  of  the 
tangent  planes  X  and  O  may  readily  be-  found  parallel  tu  those  of  the 
plane  L. 


138 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


V.     WARPED    SURFACES    OF    ROTATION. 

331.  Problem. —  TJie  axis  and  generatrix  of  a  tvarpcd  surface  of  rotation 
being  given,  to  pass  a  plane  tangent  at  a  given  point  of  the  surface. 

Let  the  surface  be  the  hyperboloid  of  one  nappe  (Fig.  277),  of  which 
{a'b',  a"b")  are  the  projections  of  the  axis,  {c'd',  c"d")  those  of  a  gen- 
eratrix, and  e"  the  horizontal  projection   of  the  given   point. 

As  the  generatrix  has  been  assumed  to  be  parallel  to  V,  the  perpendicu- 
lar a"f"  let  fall  upon  it  from  the  centre  a"  measures  the  shortest  distance 
between  the  axis  and  the  generatrix,  or,  in  other  words,  the  radius  of 
the  circle  of  the  gorge  f"g"h"  in  its  horizontal  projection.     The  vertical 

Fig.  27"r 


projection  of  the  circle  passes  through  /',  its  assumed  height  above  the 
horizontal  plane.  The  base  of  the  surface,  or  the  section  cut  by  H,  is 
determined  by  the  horizontal  piercing-point  c"  of  the  generatrix,  whose 
distance  from  a"  measures  the  radius   of  the   circle  c" I" k"  (Fig.  134). 

All  the  generatrices  must  be  projected  on  H  tangent  to  the  circle 
of  the  gorge  ;  hence  c" I"  and  e" k"  are  the  horizontal  projections  of  those 
which  pass  through  the  given  point  c.  The  vertical  projections  of  these 
generatrices  pass  through  /'  and  k'  and  the  points  of  tangency  g'  and 
//'  in  the  circle  of  the  gorge,  while  an  ordinate  from  e"  determines  the 
pomts  /  and  c' ,  the  vertical  projections  of  the  points  to  either  of  which 
a  tangent  plane   is  to   be  drawn. 

As  the   tangent    plane    must    contain  a  generatrix    and  also  a  tangent 


TANGENTS  AND   NORMALS. 


39 


line  to  the  parallel  cut  through  the  point  of  contact  (r',  c"),  and  as  the 
line  c"p"  drawn  perpendicular  to  a" e"  is  the  horizontal  projection  of 
the  tangents  to  the  two  parallels,  their  piercing-points  /'  and  /',  in  con- 
nection with  the  piercing-points  /"  and  k"  of  the  generatrices,  ar"e 
sufficient  to  fix  the  traces  of  either  tangent  plane.  Draw,  then,  the 
horizontal  traces  HM  and  HN  parallel  t<j  c" p" ,  and  let  the  vertical 
traces   pass  through   the  vertical  piercing-points  /»'  and  /'  respectively. 

A  verification  of  the  solution  will  be  found  in  the  fact  that,  the  line 
{Jig\  It-'g")  being  common  to  the  two  tangent  planes,  its  piercing-point 
d  is  also  common  to  the  two  vertical  traces. 

332.  Prohi.EM. —  Through  a  given  point  in  space  to  pass  a  plane  tangent 
to  a  ivarped  surface  of  rotation  at  a  given   meridian. 

Let  the  surface  be  given  as  in  the  preceding  case  (Fig.  278),  and  let 
{a' ,  a")  be  the  point,  and  IfL  tiie  horizontal  trace  of  the.  meridian  plane. 


As  the  tangent  plane  is  perpendicular  to  the  meridian  plane,  its  hori- 
zontal trace  is  perpendicular  to  HL.  If  now  an  auxiliary  horizontal 
plane  be  passed  through  the  given  point  {a\  a"),  it  will  cut  on  the 
tangent  plane  a  horizontal  {a"!?",  a'b')  whose  piercing-jioint  ib',  b")  is  a 
point  of  the   vertical  trace. 

This  auxiliary  plane    also   cuts  the    surface  in   a    ])arallel    whose    hori- 


140  ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 

zontal  projection  c" e"g"  may  be  found  by  means  of  the  point  of  inter- 
section {c' ,  c")  on  the  given  generatrix  (/W,  f"d"),  while  the  parallel 
itself  is  intersected  by  the  horizontal  {a"b" ,  a'b')  in  two  points  {^g" ,  g') 
and  {e",  e').  But  as  the  tangent  plane  must  contain  a  generatrix  having 
a  point  in  common  with  the  parallel  already  cut,  g"i"  and  e"k"  are  the 
projections  of  two  generatrices,  either  of  which  lies  in  a  tangent  plane 
that  satisfies  the  prescribed   conditions. 

As  all  generatrices  of  the  surface  pierce  H  in  the  circumference  of 
the  base,  the  points  h''  and  /"  are  points  of  the  horizontal  traces  of  the 
tangent  planes.  Hence,  by  leading  through  these  points  lines  parallel 
to  g"e"  the  horizontal  traces  of  the  tangent  planes  are  determined,  tb^t 
vertical  traces  of   which  pass  through   b'. 


VI.    TANGENT    SURFACES    AND    ENVELOPMENT. 

333.  The  tangency  between  surfaces  may  be  divided  into  two  general 
classes  : 

(i)  Convex  tangency,  in  which  the  convex  surfaces  touch,  whether  ii* 
point  or  line. 

(2)  Concavo-convex  tangency,  in  which  the  convex 
surface  of  one  body  is  presented  to  the  concave  surface 
of  the  other,  the  contact  being  in  a  point,  a  line  or  in  a 
series  of  separated  points. 

In  this  latter  case,  if  the  tangency  is  indicated  by  a 
closed  line — as,  for  instance,  where  one  of  the  surfaces  is- 
entirely  circumscribed  by  the  other — the  contact  is  termed 
an  envelopment. 

334.  An  enveloping  surface  may  be  generated  when  two  lines,  tan- 
gent to  each  other,  are  moved  in  such  a  way  that  the  point  of  contact 
follows  a  common  directrix. 

If  both  the  directrix  and  one  of  the  tangent  lines  be  right  lines, 
then  one  of  the  surfaces  generated  will  be  a  plane  and  the  other  a 
cylinder ;  if  both  tangents  be  curved  lines,  the  surfaces  generated  will  be 
cylindrical.  In  each  case,  however,  the  contact  will  be  in  a  line  parallel 
to  the   directrix. 

If  the  directrix  be  a  curved  line,  the  surfaces,  whatever  the  character 
of  the  tangent  lines,  will  be  curved,  the  line  of  contact  being  parallel 
to   the  directrix. 

If  the  tangent  lines  be  made  to  revolve  around  an  axis,  sui-faces  will 
be  generated  whose  line  of  contact  will  be  the  circle  described  by  the 
point  of  contact. 

335.  If  a  line  tangent  to  a  surface  be  so  moved  that  its  point  of 
contact  follows  any  plane  curve  of  the  surface,  an  enveloping  surface 
will  be  generated. 


TANGENTS  AND   NORMALS. 


141 


If  the  g-eneratrix  be  a  ri^^ht  line  moving  parallel  to  its  original  posi- 
tion, the  enveloping  surface  will  be  a  cylinder ;  but  if  one 
point  of  the  line  be  fixed,  the  enveloping  surface  will  be  a 
cone,  the  nature  of  the  line  of  contact  depending  upon  the 
character  of  the  surface  enveloped. 

336.  If  two  surfaces  be  made  tangent  to  each  other, 
and  a  portion  of  each  on  opposite  sides  of  the  line  of 
contact  be  removed,  the  remaining  sections  will  fcjrm  a 
single  surface.  The  operation  by  which  this  union  is  effected 
may  be  termed  a  tangent  juncture,  the  line  of  contact  now 
becoming  the  line  of  juncture. 

337.  The  enveloi)ing  surface  may  be  a  polyhedron  and  the  surface 
double-curved,  in  which  case  the  contact  will  be  in  separate  points, 
each  face  of  the  polyhedron  being  a  tangent  plane  to  the  surface. 

338.  Prohlem. —  To  eircuniscrihe  a  sphere  by  a  cylinder  z>.'hose  axis  is 
parallel  to  a  given  line,  ami  deter  mine  the  line  of  contact  and  the  horizontal 
trace  of  the  cylinder. 

Let  the  sphere  be  given  as  in  Fig.  279,  and  let  {a'b',  a"b")  be  the 
given  line. 


Cut  the  sphere  by  a  meridian  plane  parallel  to  the  given  line,  and  to 
the  great  circle  thus  found  lead  a  tangent  parallel  to  the  given  line. 
If,  now.  this  tangent  be  revolved  around  an  axis  which  passes  through 
the   centre  of  the  sphere  and  is   parallel   to  the   given  line,  the   point   of 


142  ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 

contact  will  describe  a  great  circle — the  line  of  contact — on  the  surface 
of  the  sphere,  and  the  tangent  will  generate  an  enveloping  cylinder 
whose  generatrices  will  be  in  the  required  position. 

Thus,  let  the  sphere  be  cut  by  the  plane  whose  horizontal  trace  HL 
is  parallel  to  a"b" ,  and  lead  through  the  centre  of  the  sphere  a  line 
{c'd' ,  c"d")  parallel  to  the  given  line.  This  is  a  line  of  the  meridian 
plane,  its  horizontal  piercing-point  d"  being  in  the  trace.  Revolve  this 
plane  around  HL  into  H,  carrying  with  it  the  great  circle  cut  on  the 
sphere  and  the  axis  {c'd',c"d")  of  the  enveloping  cylinder;  the  point  r,' 
is  the  centre  of  the  sphere,  and  d"c^  the  axis  of  the  cylinder  in  the 
new  position.  Describing,  then,  the  sphere  and  drawing  the  generatrices 
parallel  to  c^'d",    ^///  is  the  new  projection  of  the  line  of  contact. 

To  determine  the  primitive  projections  of  the  line  of  contact,  the  fol- 
lowing method  may  be  employed.  Cut  the  surface  by  a  series  of  auxiliary 
planes  which  are  parallel  to  H.  The  line  k^'g^'  is  a  section  cut  by  such 
a  plane,  the  horizontal  projection  of  which  is  a  circle  with  a  diameter 
equal  to  that  line.  The  horizontal  projection  //'  of  the  point  ///,  which 
is  common  to  the  section  and  the  line  of  contact,  is  readily  found  by 
means  of  the  ordinate,  while  the  vertical  projection  h'  may  be  deter- 
mined either  by  measuring  above  GL  the  distance  of  ///  from  the 
trace  HL,  or  by  means  of  the  auxiliary  circle  passing  through  //.'.  A 
similar  construction  for  other  points  of  the  projection  r///  determines 
\the  entire  projection  of  the  line  of  contact. 

The  horizontal  trace  or  base  of  the  enveloping  cylinder  may  be 
found  by  means  of  the  horizontal  piercing-points  of  the  various  genera- 
trices which  pass  through  the  points  of  the  line  of  contact  above 
determined.  Thus,  the  generatrices  which  contain  the  points  (///,  h") 
are  parallel  to  the  axis  of  the  enveloping  C3dinder,  and  pierce  H  in  the 
points  in"  and   m"  of  the  base  sought. 

339.  Problem. —  Through  a  given  point  in  space  to  pass  a  plane  tangent  to 
a  surface  of  revolution,  the  point  of  contact  being  on  a  given  parallel. 

Let  the  surface  be  given  as  in  Fig.  280,  and  let  {b'c' ,  b"c")  be  the 
parallel,  and  {a' ,  a")  the  point. 

Imagine  a  tangent  to  be  drawn  to  any  meridian  at  a  point  of  the 
given  parallel,  and  to  be  revolved  around  the  surface  with  its  point 
of  contact  constantly  in  that  parallel  ;  it  will  generate  an  enveloping 
cone  whose  vertex  is  in  the  axis.  Any  plane  tangent  to  this  cone  will 
be  tangent  to  the  surface  enveloped  at  a  point  of  the  line  of  contact 
or  parallel.     Hence  the  problem  is  resolved  into  Art.  325. 

Should  the  vertex  fall  outside  the  limits  of  the  drawing,  the  follow- 
ing method  may  be  employed :  Draw  the  tangent  {b' d\  b"d")  to  the 
principal  meridian  ;  this  is  a  generatrix  of  the  enveloping  cone,  the  base 
of  which  is  the  distance  of  d"  from  the  foot  of  the  axis.  Through  the 
given  point  pass  a  secant  plane  cutting  a  circle  on  the  auxiliary  cone, 
and  draw  the  tangents  a"ni"  and  a"n''  to  that  circle;  these  hues  are 
horizontals  of  the   required   tangent   planes,  and   their  tangent   points   m" 


TANGENTS  AND  NORMALS. 


143 


and  n"  points  in  the  generatrices  of  the  cone.  Hence  o"g"  and  o"h" 
are  the  horizontal  projections  of  the  lines  of  contact,  their  intersections 
(/',  e')  and  (/",  /')  with  the  given  parallel  being  the  points  of  contact 
of  the  tangent  planes  with  the  given  surface. 


The  points  //"  and  _i^".  the  feet  of  the  generatrices,  are  the  points  in 
which  the  traces,  IIL  and  //J/,  are  tangent  to  the  base  of  the  envelop- 
ing cone  ;  and  as  the  lines  am  and  an  are  lines  of  the  tangent  planes, 
their  piercing-points  {k\  k")  and  (/',  /")  detcnninc  the  vertical  traces  fX 
and    I'M  respectively. 

340.  Prohlkm. —  To  determine  a  similar  tangencv  by  means  of  an  auxiliary 
Sphere. 

Let  the  surface  be  given  as  in  Fig.  281,  and  let  ^b'e\  b"e")  be  the 
parallel,  and  (n',  a")  the    point. 

Draw  a  tangent  {b'el' ,  b"d")  to  the  principal  meridian,  and  a  normal 
{b'e\b"e")  at  the  point  of  contact;  b'e'  is  the  true  length  of  the  radius 
of  the  sphere,  which  is  enveloped  by  the  given  surface  of  revolution, 
the  line  of  juncture  being  the  given   parallel. 

Through  the  given  point  imagine  a  secant  plane  to  be  led,  cutting 
a  meridian  section  on  the  sphere,  and  further  imagine  two  tangents  to 
be  drawn  to  the  great  circle  so  cut.  If.  now.  the  tangents  be  revolved 
around  the  line  {a'e\  a"e")  as  an  axis,  thev  will  describe  an  enveloping 
cone  whose   hne  of*  juncture  is  the   locus  of    the   point  of  contact.     Any 


144 


ELEMENTS  OF  DESCRIPTIVE    GEOMETRY. 


plane  which  passes  through  («',  a")  and  is  tangent  to  the  cone  will  be 
tangent  to  the  sphere ;  but  to  be  tangent  to  the  given  surface  at  the 
same  time  it  must  contain  the  point  of  intersection  between  the  line  of 
juncture  and  the  given  parallel. 

To  effect  this  construction  rotate  the  plane  Z,  carrying  with  it  the 
tangents  and  the  meridian  section,  around  the  axis  of  the  given  surface, 
until  it  assumes  a  position  parallel  to  F;  the  trace  HL  then  coincides 
with  c"b"  and  the  great  circle  with  the  apparent  contour  b'/^e'  of  the 
sphere,  while  the  given  point  falls  at  («/',  a^),  and  the  vertical  projec- 
tions of  the  tangents  at  a^g^  and  «///.  The  line  ^///  is  the  vertical 
projection  of  the  revolved   line  of  juncture  or  circle  of  contact  betweeii 


Fig.  281 


the  auxiliary  cone  and  sphere,  and  its  intersection  ///  with  the  given 
parallel  is  the  revolved  point  of  contact  between  the  required  tangent 
plane  and  the  given  surface. 

-Determine  ///'  and  h^' ;  then  by  counter-rotation  of  the  plane  L  the 
line  h,"h,",  which  retains  its  perpendicular  position  to  that  plane,  is  pro- 
jected in  h"h",  and  marks  by  its  extremities  the  horizontal  projections  of 
the   points  of  contact   between  the  tangent  planes  and  the  given  surface.- 

As  the  tangent  plane  is  perpendicular  to  the  normal  {c'h',  c"h")  at  the 
point  of  incidence,  the  traces  may  be  determined  by  passing  through 
the  given  point  a  horizontal  {a''p\  a"p")  of  that  plane  ;  its  piercing-point 
{P'^P")^  in  connection  with  the  normal,  is  sufficient  to  fix  the  position  of 
the  tangent  plane  required. 


DE  VE  L  OPMEX  T.  14^ 


CHAPTER   XIV. 

DEVELOPMENT. 

341.  If  a  surface  be  rolled  upon  a  plane  so  as  to  brinj^  each  con- 
secutive face  in  polyhedrons,  or  each  consecutive  right  line  in  ruled 
surfaces  in  contact  with  that  plane  without  stretching,  folding  or  tear- 
ing the  face  or  element,  the  surface  is  said  to  be  developed,  and  the  plane 
figure  which   results  therefrom  is  termed  its  development. 

342.  Rectilinear  Surfaces. — All  polyhedrons  are  developable,  inasmuch 
as  each  face,  being  a  plane  figure,  may  be  brought  in  regular  succession 
in  contact  with  a  plane,  the  adjacent  faces,  two  and  two,  having  an 
edge  in  common.  As  all  developed  surfaces  necessarily  represent  the 
various  parts  of  an  object  in  their  true  dimensions,  these  must  be  deter- 
mined from  the  projections  either  by  direct  calculation  or  by  the  graph- 
ical methods  already  explained. 

343.  Prismatic  surfaces  in  which  the  edges  are  parallel  present  de- 
velopments in   which   the  edges  are  parallel. 

If  the  surface  be  limited  by  bases  at  right  angles 
to  the  axis  or  edges,  the  perimeter  of  the  bases  will 
be  developed  into  ri^ht  lines  at  right  angles  to  the 
axis,  the  length  of  the  same  being  equal  to  the  sum 
of  the  sides  of  the  bases. 

With  oblique  prisms  the  bases  incline  to  the  axes,  hence  have  sides 
which,  as  a  general  rule,  incline  to  the  edges  and  will  always  develop 
into  broken  lines. 

344.  Pyramidal    surfaces    in    which    the    edges    converge    to    a    vertex 
present    developments    in    which    the    edges    con- 
verge, forming,  two  and  two,  the  boundary  lines 
of  the  plane  faces. 

Should  the  surface  be  limited   by  a  base,  each 
face  will  be  developed  as  a  triangle,  the  distance 
of  the  angular   points  of  the  base   from   the   ver- 
tex being  measured   by  the   lengths  of   the   respective   edges  of   the    sur- 
face and   the   base   unrolling   in  a  broken  line. 

With  the  right  regular  pyramid,  the  edges  being  equal,  the  base 
will  be  contained  within  the  circumference  of  a  circle  whose  radius  is 
equal  to  the  length  of  an  edge. 

345.  All  points,  lines  and  sections  of  a  polyhedral  surface  preserve, 
after  development,  their  relative  positions  on  the  separate  faces. 


146 


ELEMENTS   OF  DESCRIPI'IVE    GEOMETRY. 


346.  Ruled  Surfaces. — All  surfaces  in  which  the  consecutive  right  ele- 
ments are  lines  in  the  same  plane  are  capable  uf 
development.     AH  others  are  undevelopable. 

If   a   cylindrical   surface   be   rolled   on   a   plane 
until    it    returns    to    the    element    from    which    it 
started,  the  surface  of  the  plane  which  its  succes- 
sive   elements    have    covered  will   be   the    measure 
of  the  exterior  surface  and,  hence,  its  development. 
If  the   surface  be  limited   by  bases  at  right   angles   to   the   axis,  they 
will  during  the  unrolling  maintain  that  position  to  the  plane  of  develop- 
ment,   and  hence  will    be    developed  in  the   traces  of   their  planes,  or  as 
right  lines  perpendicular  to  the  elements. 

Should  the  bases  be  other  sections  of  the  surface,  their  developments 
will  be  curved  lines,  each  point  of  which  will  be  determined  by  the 
extremities  of  the  elements  cut  by  the  secant  plane. 

347.  In  bringing  the  successive  right  elements  of  a  conic  surface  in 
contact  with  a  plane,  the  vertex  remains  fixed, 
and  the  extremities  of  the  elements,  should  the 
surface  be  limited  by  a  base,  will  mark  the  suc- 
cessive points  of  that  base. 

With  the  right  cone  of  revolution,  the  elements 
being  equal,  the  points  of  the  base  are  at  the 
same  distance  from  the  vertex,  and  hence,  when  developed,  will  lie  in 
the  arc  of  a  circle  whose  radius  is  equal  to  the  length  of  an  element. 

348.  Tangents  and  normals  to  curves  will  be  tangents  and  normals 
to  their  developments  at  the  points  of  contact  and  incidence  respectively. 

349.  Problem. —  To  develop  a  right  prism  with  oblique  section. 
Let  the  surface   be    given  as  in  Fig.   282. 

Fig.  38S 


c: 

\ — ^ 

— 

d 

6 

^- 

-- 

d 

-6 

f 

c 

e' 

d' 

Since  the  base  oi  the  prism  will  be  developed  as  a  right  line,  lay 
off  upon  an  indefinite  line,  AB,  the  sides  of  that  base,  given  in  their 
true  size  in  a'd",    b"c",  c"d",   etc.,  and    erect  at  the    points  a,    b,   c,    etc.. 


DEVELOPMENT. 


147 


thus  found  a  series  of  perpendiculars.  These  represent  the  edges  of  the 
surface,  and  arc  equal  to  the  length  as  determined  in  the  vertical  pro- 
icction. 

Upon  these  lines  measure  the  distances  aa,  bb,  cc,  etc.,  equal  to  those 
jortions  of  the  edges  which  are  below  the  secant  plane,  and  are  indi- 
cated  on  the  vertical  plane  in  their  true  lengths  a'a\  b'b',  c'c',  etc.  The 
broken  line  thus  determined  is  the  development  of  the  section. 

350.    Problem. —  To  develop  the  section  of  ei  right  cylinder. 

By  considering  the  surface  as  a  prism,  whose  base  is  a  polygon  of 
an  infinite  number  of  sides,  it  is  evident  that  the  principles  involved  in 
its  development  are  precisely  those  employed  in  the  preceding  case. 

Let  the  surface  be  given  as  in  Fig.  283. 

Divide  the  base  into  any  number  of  small  arcs,  the  length  of  the 
chord  being  approximately  equal  to  the  arc  which  it  subtends.  Upon 
an  indertnite  line  lay  off  these  chords,  measuring  the  perimeter  of  the 
base  in  the  line  hh.  Erect  at  the  points  a,  b,  c,  etc.,  the  perpendiculars 
representing  the  elements  of  the  surface  which  pass  through  the  extremi- 
ties of   the  chords,  and   lay   off   upon  them    the   distances   aa,  bb,  cc^  etc., 

Kig.  aB3 


equal  to  the  true  lengths  of  those  elements  as  projected  on  the  vertical 
plane.  The  curve  passing  through  extremities  of  the  perpendiculars  thus 
found  is  the  development  of  the  section. 

To  drazv  a  tangent  to  the  developed  curve  at  any  point  t.^Lay  otT  the 
distance  of  the  subtangent  ci  equal  to  c"i"\  join  c  of  the  developed 
curve  and  i  for  the  required  tangent,  since  the  plane  of  the  drawing  is 
tangent  at  the  right  element  passing  through  c,  and  hence  contains  both 
tangent  and   subtangent. 

351.    Prohlem.—  To  develop  the  section  of  an  oblique  cylinder. 

Let  the  cylinder  be   given  as  in   Fig.  284. 

Since  the  section  perpendicular  to  the  axis  of  a  prism  or  a  cylinder 
will  always  develop  into  the  right  line.  tind.  by  Fig.  246.  the  projections 
and  true  size  of  such  a  section  the  jierimeter  of  which  measures  the 
width  of  the  developed    surface.     Then  upon    an   indefinite    line,  AB,  lay 


148 


ELEMENTS   OF  DESCRIPTIVE   GEOMETRY. 


off  the  distances  ab,  be,  cd,  etc.,  equal  to  a,"b^",  b,"c^",  r/W/',  etc.,  the 
chords  of  the  ellipse  which  are  approximately  equal  to  the  arcs  which  they 
subtend. 

Kig.  S84 


To  determine  the  distances  of  the  points  of  the  base  from  the.  cross- 
section,  rotate  any  element  as  {e"e",  eV)  parallel  to  V\  eV/  is  its  vertical 
projection,  and  measures  with  GL  the  inclination  of  all  the  elements  to 
H.  During  rotation  the  point  on  any  one  element  does  not  alter  its 
distance  from  //;  hence  horizontal  lines  drawn  from  the  vertical  projec- 
tions of  the  different  points  of  the  section,  a',  b' ,  c' ,  etc.,  determine  the 
required  distances.  Thus  the  length  of  the  element  below  the  point 
{a  ,  a")  is  a,'e',  below  {b' ,  b")  is  b^e' ,  etc.  Laying  off  these  distances  on 
perpendiculars  to  the  developed  section  drawn  through  a,  b,  c,  etc.,  the 
curve  passing  through  their  extremities  is  the  development  of  the  base. 

To  draw  a  tangent  to  the  developed  curve  of  the  base  at  any  point  fi. 
— The  line  will  be  the  developed  position  of  o" ft"  tangent  to  the  base, 
of  which  o"b"  tangent  to  the  section  is  the  subtangent ;  hence,  making 
bo  in  the  development  equal  to  b"o" ,  the  tangent  to  the  section,  the 
line  joining  0  and  ft  is  the  required  tangent  to  the  developed  curve  ol 
the  base. 

352.    Problem. —  To  develop  a  right  regular  pyramid  and  section. 

Let  the  surface  be  given  as  in  Fig.  285. 

The  edges  being  equal,  the  angular  points  of  the  base  lie  in  an  arc 
of  a  circle   whose  radius  is  equal  to  the  length   of  an   edge. 

Determme  this  length  by  rotating  an  edge,  as  {7>'d\  7<"d"),  parallel  to  T. 
With  v'd^'  as  a  radius  describe  an  indefinite  arc,  the  centre  of  which^  is 
the  vertex  in  the  developed  surface. 

Assume  an  edge  %>a,  and  from  a  lay  off  the  chords  ab,  be,  cd,  etc., 
respectively  equal    to  the  sides  of   the    base   a"b",  b"c",  c"d",  etc. ;    from 


DEVELOPMENT. 


149 


the  points  a,  b,  c,  etc.,  thus  found    draw   the   edges  to   v,  and   the   figure 
determined  is  the  development  sought. 

To  develop  the  section,  draw  through   the   breaking-points  thereof    hori- 


Kig.  28o 


zontal  lines  intersecting  the  revolved  edge  v'd^\  the  distances  from  :''  to 
the  points  thus  determined,  when  measured  from  v  upon  the  developed 
edges,  mark  the  points  through   which  the  developed  section  passes. 

353,    Problem. —  To  develop  a  right  cone  of  revolution  and  siction. 

Let  the  surface  be  given  as  in  Fig.  286. 


Regarding  the  cone  as  a  pvraniid  whose  base  is  a  |>»>i\gon  01  an 
infinite  number  of  sides,  the  princii)lcs  involved  are  identical  with  those 
of  the  preceding  case. 

Divide  the  circumference  of  the  base  into  small  arcs  whose  chords 
may  be  taken  as  equal  to  those  arcs;  describe  by  means  of  an  element 
an   indefinite  arc.  and   lav  off    unon    it   the    successive  chords  of  the    base. 


I50 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


The  developed  surface  is  the  sector  ot  a  circle,  in  which  the  centre  is 
the  vertex,  the  radii  the  terminal  elements  of  the  developed  surface,  and 
the  arc  of  the  circle  the  measure  of  the  perimeter  of  the  base. 

To  develop  the  section,  lead  through  the  points  in  which  the  secant 
plane  cuts  the  elements  passing  through  the  dividing  points  of  the  base, 
horizontal  lines  which  intersect  the  element  iy' e' ,  v" e")  ;  the  distances 
from  v  to  the  points  thus  determined,  when  measured  from  v  upon  the 
developed  elements,  mark  the  points  through  which  the  developed  section 
passes. 

The  tangent  to  any  point,  as  /,  of  the  developed  section  is  found  by- 
making  bo  equal  to  the  subtangent  b" o" . 

354.    Problem. —  To  develop  the  oblique  pyj'amid  and  section. 

Let  the  surface  be  given  as  in  Fig.  287. 


The  triangular  faces  when  developed  must  be  exhibited  in  their  full 
size;  hence,  determine  the  true  lengths  of  the  edges  by  rotation.  The 
measurements  thus  f6und,  in  connection  with  the  base  lines,  already  hori- 
zontally projected  in  their  true  lengths,  determine  the  development. 

Thus,  let  it  be  required  to  find  the  face  {v'a'b',  v"a"b"). 

Rotate  the  edges  {v'a',  v"a")  and  {v'b',  v"b")  around  the  vertical  line 
passing  through  the  vertex  until  they  assume  positions  parallel  to  V  \ 
v"a"  and  v"b"  fall  respectively  in  v"al'  and  v"bl',  and  v'a^  and  v'b^  are 
the  vertical  projections  in  the  new  position.  With  v'a^  and  -o'b'  as  radii, 
describe  indefinite  arcs,  and  draw  at  will  the  developed  edge  v'a.  From 
«  as  a  centre,  and  with  a  radius  equal  to  the  side  of  the  base  of  the 
pyramid,  describe  an  arc;  the  point  of  intersection  b  between  this  arc 
and  that  described  through  the  extremity  b'  completes  the  face  vab 
required. 

In  a  similar  manner  the  remaining  faces  may  be  readily   determmed. 

As  the  breaking-points  of  the  section  do  not  alter  their  heights  above 
H  during    the    rotation    of   the  edges,   horizontal    lines    drawn    from    their 


DEVELOPMENT. 


15' 


vertical  projections  mark  at  their  points  of  intersection  with  the  revolved 
edges  their  true  distances  from  the  vertex. 

355.    Problem. —  To  develop  the  oblique  cone. 

Let  the  surface  be  given  as  in  Fig.  288. 

By  regardmg  the  cone  as  an  obHque  pyramid,  having  for  its  base  a 
polygon  of  an  mfinite  number  of  sides,  the  solution  will  be  iacnti^:ii 
with  the  precedmg  case. 

An  elegant  solution  may  likewise  be  effected  by  means  of  the  inter- 
secting sphere,  as  follows: 

From  the  vertex  of  the  cone  as  a  centre  describe  any  sphere  inter- 
secting the  surface  (jf  the  cone  m  a  line  all  of  whose  points  arc  necessarily 


d'i    ~    c'{ 


b'i    a" 


at  the  same  distance  from  the  vertex.  Determine  this  line  of  intersection 
by  passing  through  the  cone  a  series  of  vertical  secant  planes  which  cut 
right  elements  on  that  surface  and  great  circles  on  the  sphere,  and 
rotate  the  planes  and  sections  parallel  to  V  around  a  vertical  axis  pass- 
ing through  {i'\  v").  Thus,  the  element  {z'"a'\  v'a')  falls  at  {v"a,'\  v'a,'\ 
the  section  of  the  sphere,  in  each  case,  coinciding  with  the  primitive  vertical 
projection.  Drawing,  then,  a  horizontal  line  through  <r,',  its  intersection 
with  the  element  passing  through  ui'.  n")  of  the  base  determines  a  point 
a'  of  the  required  line.  By  a  similar  procedure  the  remaining  points 
necessary  to  complete  the  entire  line  of  intersection  may  be  found. 

As  the   line  of  intersection  is  not  a   plane  curve,  it  cannot  be   rotated 


1^2 


ELEMENTS   OF  DESCRIPTIVE    GEOMETRY. 


parallel  to  either  plane  of  projection;  the  true  length  of  the  perimeter 
may,  however,  be  obtained  by  regarding  the  horizontal  projection 
a'  ft"y"S"  as  the   base  of  a  right  cylinder. 

Develop  this  imaginary  cylinder;  its  base  is  the  right  line  a^"  .  .  . 
6,"  .  .  .  a/',  from  which  the  development  of  the  line  of  intersection  is 
effected  by  laying  off  the  heights  of  its  respective  points  above  H. 

Again,  as  the  line  of  intersection  has  each  point  at  the  same  distance 
from  the  vertex,  describe  an  indefinite  arc  with  a  radius  equal  to  that 
of  the  auxiliary  sphere,  and  measure  upon  this  arc  the  distances  AB, 
BC,  .  .  .  equal  to  ^^/Z^/,  /^/y/  ...  of  the  developed  line  of  intersection. 
Through  the  points  A,  B,  C,  etc.,  thus  determined  draw  the  elements 
of  the  cone,  and  lay  off  below  the  developed  line  of  intersection  the 
additional  lengths  of  the  elements  necessary  to  complete  the  conic  sur- 
face. Thus,  from  A  lay  off  the  distance  Aa  equal  to  a/  a/,  found  in 
its  true  size  on  the  element  in  its  revolved  position,  from  B  the  distance 
/y/^/,  and  so  on   for  the  remaining   points  of  the  base. 

356.    Problem. —  To  develop  the  icosahedron. 

Let  the  surface   be  given   as  in   Fig.  289. 


An  inspection  of  the  surface  shows  that  the  middle  zone  is  composed 
of  a  series  of  ten  equal  triangles,  the  length  of  the  sides  being  equal  to 
the  edge  a'b'  projected  on  V  in  its  true  size,  and  the  base-lines  forming 
two  regular  pentagons  parallel  to  H.  By  development  these  two  poly- 
gons  become   two   parallel    lines,  between    which    lie   the    ten    equilateral 

triangles. 

As  the  polygons  are  likewise  the  bases  of  two  pyramids  whose  faces 
are  equilateral'  triangles,  develop  the  pyramids,  keeping  one  side  of  each 
base  in  common  with   the  development  of  the   zone. 


,-5^  "x* 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 
LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 
This  book  IS  du^^  ^^^^  ^^  ^_^.^^  ^^^^^  d. 

Renewed  books  are  subjea  toimmedia^^ 


'itrm'^f-l' 


^P0 


REC'D  LD 


FEB  8  1958 


LD  21A-50m-8/57 
(C8481sl0)476B 


General  Library     . 
University  of  California 
Berkeley 


t 


y 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


